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码题集新手村600道(前300道)
码题集新手村600道[不含vip]
- 刷题链接
- MT1001
- MT1002
- MT1003
- MT1006
- MT1007
- MT1008
- MT1009
- MT1010
- MT1011
- MT1012
- MT1013
- MT1015
- MT1016
- MT1017
- MT1018
- MT1020
- MT1021
- MT1022
- MT1024
- MT1025
- MT1026
- MT1027
- MT1033
- MT1036
- MT1037
- MT1041
- MT1042
- MT1043
- MT1044
- MT1045
- MT1046
- MT1047
- MT1049
- MT1050
- MT1051
- MT1052
- MT1053
- MT1054
- MT1055
- MT1056
- MT1057
- MT1058
- MT1059
- MT1060
- MT1061
- MT1063
- MT1064
- MT1065
- MT1066
- MT1067
- MT1068
- MT1069
- MT1071
- MT1073
- MT1074
- MT1075
- MT1076
- MT1077
- MT1079
- MT1080
- MT1081
- MT1082
- MT1083
- MT1085
- MT1086
- MT1087
- MT1088
- MT1090
- MT1096
- MT1097
- MT1100
- MT1101
- MT1102
- MT1103
- MT1105
- MT1107
- MT1108
- MT1109
- MT1110
- MT1111
- MT1112
- MT1114
- MT1115
- MT1116
- MT1117
- MT1119
- MT1121
- MT1122
- MT1123
- MT1124
- MT1127
- MT1128
- MT1129
- MT1130
- MT1131
- MT1132
- MT1133
- MT1134
- MT1138
- MT1139
- MT1140
- MT1141
- MT1142
- MT1144
- MT1146
- MT1148
- MT1149
- MT1150
- MT1151
- MT1152
- MT1153
- MT1154
- MT1155
- MT1157
- MT1159
- MT1161
- MT1163
- MT1167
- MT1168
- MT1170
- MT1171
- MT1172
- MT1173
- MT1174
- MT1176
- MT1178
- MT1183
- MT1184
- MT1185
- MT1186
- MT1187
- MT1188
- MT1189
- MT1190
- MT1191
- MT1192
- MT1193
- MT1194
- MT1195
- MT1196
- MT1197
- MT1198
- MT1199
- MT1201
- MT1203
- MT1204
- MT1206
- MT1207
- MT1208
- MT1211
- MT1212
- MT1213
- MT1214
- MT1215
- MT1216
- MT1217
- MT1219
- MT1220
- MT1222
- MT1223
- MT1224
- MT1226
- MT1228
- MT1230
- MT1232
- MT1233
- MT1234
- MT1236
- MT1239
- MT1241
- MT1242
- MT1244
- MT1246
- MT1247
- MT1249
- MT1250
- MT1252
- MT1254
- MT1255
- MT1256
- MT1257
- MT1260
- MT1262
- MT1263
- MT1264
- MT1265
- MT1266
- MT1267
- MT1268
- MT1269
- MT1270
- MT1272
- MT1273
- MT1274
- MT1275
- MT1276
- MT1277
- MT1278
- MT1280
- MT1282
- MT1285
- MT1286
- MT1287
- MT1289
- MT1290
- MT1291
- MT1294
- MT1295
- MT1298
- MT1299
刷题链接
MT1001
#include<bits/stdc++.h> using namespace std;int main( )
{cout<<"This is my first program!"<<endl<<
"Coding is fun!";return 0;
}
MT1002
#include<bits/stdc++.h> using namespace std;
int num;
int main( )
{cin>>num;cout<<"You entered:"<<num;return 0;
}
MT1003
#include<bits/stdc++.h> using namespace std;
int x,y;
int main( )
{scanf("%d,%d",&x,&y);cout<< x << "+" << y << "=" <<x+y<<endl;cout<< x << "-" << y << "=" <<x-y<<endl;return 0;
}
MT1006
#include<bits/stdc++.h> using namespace std;
double x,y;
int main( )
{cin>>x>>y;//printf("%0.6f\n",x);//printf("%0.6f\n",y);printf("%0.6f*%0.6f=%0.6f\n",x,y,x*y);printf("%0.6f/%0.6f=%0.6f",x,y,x/y);return 0;
}
MT1007
#include<bits/stdc++.h> using namespace std;
double x,y,z;
int main( )
{cin>>x>>y>>z;//cout<<x <<" "<<y <<" " <<z<<endl;printf("%0.6f\n",(x+y+z));printf("%0.6f",(x+y+z)/3);return 0;
}
MT1008
#include<bits/stdc++.h> using namespace std;
const double PI = 3.1415926;
double r,h;
int main( )
{cin >> r >> h;//cout << r << " " << h << endl;printf("%0.2f\n",(PI*2*r));printf("%0.2f\n",(PI*r*r));printf("%0.2f\n",(PI*r*r*4));printf("%0.2f\n",(PI*r*r*r*(4.0/3)));printf("%0.2f\n",(PI*r*r*h));return 0;
}
MT1009
#include<bits/stdc++.h> using namespace std;
int x,y;
int main( )
{cin >> x >> y;//cout<< x <<" "<< y;printf( "%0.2f\n", (1/2.0) * ( y*x*1.0 + (y+x)/(4.0*y) ) );return 0;
}
MT1010
#include<bits/stdc++.h> using namespace std;
char x,y;
int main( )
{scanf("%c,%c",&x,&y);//printf("%c %c\n",x,y);//printf("%d %d\n",x,y);printf("The ASCII code of %c is %d\n",x,x);printf("The ASCII code of %c is %d\n",y,y);return 0;
}
MT1011
#include<bits/stdc++.h> using namespace std;int main( )
{char a = 'X';int b = 65;cout<<a<<" "<<int(a)<<endl;cout<<char(b)<<" "<<b<<endl;return 0;
}
MT1012
#include<bits/stdc++.h> using namespace std;int main( )
{cout<<"Size of int: "<<sizeof(int)<<" bytes"<<endl<<"Size of float: "<<sizeof(float)<<" bytes"<<endl <<"Size of double: "<<sizeof(double)<<" bytes"<<endl <<"Size of char: "<<sizeof(char)<<" byte"<<endl;return 0;
}
MT1013
#include<bits/stdc++.h> using namespace std;int main( )
{cout<<"Size of int = "<<sizeof(int)<<" bytes"<<endl<<"Size of long int = "<<sizeof(long int)<<" bytes"<<endl <<"Size of long long int = "<<sizeof(long long int)<<" bytes"<<endl <<"Size of double = "<<sizeof(double)<<" bytes"<<endl<<"Size of long double = "<<sizeof(long double)<<" bytes"<<endl;return 0;
}
MT1015
#include<bits/stdc++.h> using namespace std;int main( )
{int a,c;char b,d;scanf("a=%d,b=%c,c=%d,d=%c",&a,&b,&c,&d);printf("%d %c %d %c",a,b,c,d);return 0;
}
MT1016
#include<bits/stdc++.h> using namespace std;//宽度为5,%+5d,右对齐
//%-5d,左对齐
/*
左对齐 右对齐
_____ _____
455 | 455
-123 | -123
987654| 987654
*/
int main( )
{printf("%-5d %5d\n",455,455);printf("%-5d %5d\n",-123,-123);printf("%-5d %5d\n",987654,987654);return 0;
}
MT1017
#include<bits/stdc++.h>
//%m.nf
//m代表宽度,n代表精度using namespace std;
/*
3.141593 3.141593
22.345600 22.345600
*/
int main( )
{printf("%-14.6f %14.6f\n",3.141593,3.141593);printf("%-14.6f %14.6f\n",22.345600,22.345600);return 0;
}
MT1018
#include<bits/stdc++.h> using namespace std;
/*
123456789
*/
/*
123 456 789
*/
//变量a1,a2,a3
int main( )
{int a1,a2,a3;scanf("%3d%3d%3d",&a1,&a2,&a3);printf("%d %d %d",a1,a2,a3);return 0;
}
MT1020
#include<bits/stdc++.h> using namespace std;
//%m.ns
//m代表宽度,n代表精度 //小数点前面是宽度,小数点后面是精度(常用)
//m=6,表示占6列
//n=3代表输出前3个字符
int main( )
{char c[100];//scanf("%s",c);/*gets()cin.getline()getline()*///gets(c); 可以用来输入一行cin.getline(c,sizeof(c));printf("%6.3s",c);//printf("\n%.3d",6);//006return 0;
}
MT1021
#include<bits/stdc++.h> using namespace std;
/*
31331.14345435
*/
/*
31331.143454 31331.14 31331.143
*/
int main( )
{/*float型输入用%fdouble型输入用%lf*/double n;scanf("%lf",&n);printf("%lf %.2lf %8.3lf",n,n,n);return 0;
}
MT1022
#include<bits/stdc++.h> using namespace std;/*
3.1415926、12345678.123456789
*/
/*
3.141593 3.141593e+000
12345678.123457 1.234568e+007
*/
void test(double x){printf("%lf ",x);int tmp=0;while(x>=10){tmp++;x = x/10;}printf("%lfe+%.3d",x,tmp);
}
int main( )
{//printf("3.141593 3.141593e+000\n12345678.123457 1.234568e+007");//printf("%lf %e\n",3.141593,3.141593);//printf("%lf %e",12345678.123457,12345678.123457);test(3.141593);printf("\n");test(12345678.123457);return 0;
}
MT1024
MT1025
#include<bits/stdc++.h> using namespace std;/*
%o 八进制 以0开头:%#o
%x等价于%X 十六进制 以0开头:%#x 或者%#X
*/
int main( )
{/*202、117、70、130*/printf("%d %#o %#X\n",202,202,202);printf("%d %#o %#X\n",117,117,117);printf("%d %#o %#X\n",70,70,70);printf("%d %#o %#X\n",130,130,130);return 0;
}
MT1026
#include<bits/stdc++.h> using namespace std;int main( )
{int a,b,c;scanf("%d %d %d",&a,&b,&c);printf("%d",(100*a+10*b+c));return 0;
}
MT1027
#include<bits/stdc++.h> using namespace std;int main( )
{//两位数int a;scanf("%d",&a);//printf("%d%d\n",(a%10),(a/10));printf("%.2d",(a/10+(a%10)*10));//必须要写成%.2d,因为有特例10/*int a,b,c;scanf("%d",&a);b=a/10;c=a%10;a=b+c*10;printf("%.2d",a);*/return 0;
}
MT1033
#include<bits/stdc++.h> using namespace std;int main( )
{//pow(x,y);x的y次方int x;cin>>x;cout<<pow(x,6)-2*pow(x,5)+3*pow(x,4)-5*pow(x,2)+6*x+7<<endl;return 0;
}
MT1036
#include<bits/stdc++.h> using namespace std;int main( )
{int x,y;cin>>x>>y;//cout<<pow((x+y),3);//运行出错cout<<(x+y)*(x+y)*(x+y);//通过return 0;
}
MT1037
#include<bits/stdc++.h> using namespace std;int main( )
{double a;cin>>a;//if(a<0)a=-a;//else if(a>0)a=a;//else if(a==0)a=0;printf("%lf",abs(a));return 0;
}
MT1041
#include<bits/stdc++.h> using namespace std;
const double n=3.1415926;
//s=π*r*r
//c=2*π*r
int main( )
{double r;cin>>r;printf("Area=%.6lf\n",n*r*r);printf("Circumference=%.6lf\n",2*n*r);return 0;
}
MT1042
#include<bits/stdc++.h> using namespace std;
const double n=3.1415926;int main( )
{double c,k;cin>>c>>k;printf("Area=%.6lf\n",c*k);printf("Perimeter=%.6lf\n",2*(c+k));return 0;
}
MT1043
#include<bits/stdc++.h> using namespace std;
const double n=3.1415926;int main( )
{double c,k;cin>>c>>k;printf("Area = %.6lf\n",n*c*k);return 0;
}
MT1044
#include<bits/stdc++.h> using namespace std;
const double n=3.1415926;int main( )
{double c,k;cin>>c>>k;printf("Area=%.2lf\n",c*k*0.5);return 0;
}
MT1045
#include<bits/stdc++.h> using namespace std;
const double n=3.1415926;int main( )
{double c,k;cin>>c>>k;printf("%lf\n",c*k);return 0;
}
MT1046
#include<bits/stdc++.h> using namespace std;
const double n=3.1415926;int main( )
{double c,k;cin>>c>>k;printf("%.2lf\n",c*k*0.5);return 0;
}
MT1047
#include<bits/stdc++.h> using namespace std;
const double n=3.1415926;int main( )
{double x,y,z;cin>>x>>y>>z;printf("%.2lf\n",(x+y)*z*0.5);return 0;
}
MT1049
#include<bits/stdc++.h> using namespace std;int main( )
{double x1,y1,x2,y2,x3,y3;cin>>x1>>y1>>x2>>y2>>x3>>y3;double S;S=0.5*abs(x1*y2+x2*y3+x3*y1-x1*y3-x2*y1-x3*y2);printf("%.2lf",S);return 0;
}
MT1050
#include<bits/stdc++.h> using namespace std;int main( )
{double x1,y1,z1,x2,y2,z2,x3,y3,z3;cin>>x1>>y1>>z1>>x2>>y2>>z2>>x3>>y3>>z3;double S;double a,b,c;a=sqrt( pow(x1-x2,2)+pow(y1-y2,2)+pow(z1-z2,2) );b=sqrt( pow(x1-x3,2)+pow(y1-y3,2)+pow(z1-z3,2) ); c=sqrt( pow(x2-x3,2)+pow(y2-y3,2)+pow(z2-z3,2) );double p;p=1.0/2*(a+b+c);//海伦公式:s=sqrt(p*(p-a)*(p-b)*(p-c));//p=1.0/2*(a+b+c);S=sqrt(p*(p-a)*(p-b)*(p-c));printf("%.2lf",S);return 0;
}
MT1051
#include<bits/stdc++.h> using namespace std;//题目解析:拆成4个三角形
/*
/?t=553.3
*/double test
(double x1,
double y1,
double x2,
double y2,
double x3,
double y3)
{//12 13 //23 21//31 32/*怎么记这个公式???, 13 完了后面跟着2321 完了后面跟着31*/return 1.0/2*abs(x1*y2-x1*y3+x2*y3-x2*y1+x3*y1-x3*y2);
}int main( )
{double x1,y1,x2,y2,x3,y3,x4,y4;cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;//分解成4个三角形,三角形的面积公式://S=1/2*|x1y2-x1y3+x2y3-x2y1+x3y1-x3y2|;//算出来是一个三角形,所以要x2double S1,S2,S3,S4;S1=0.5*test(x1,y1,x2,y2,x3,y3);//123S2=0.5*test(x1,y1,x2,y2,x4,y4);//124S3=0.5*test(x1,y1,x3,y3,x4,y4);//134S4=0.5*test(x2,y2,x3,y3,x4,y4);//234printf("%.2lf",S1+S2+S3+S4);return 0;
}
MT1052
#include<bits/stdc++.h> using namespace std;
//angle 角度
//radian 弧度
const double pai=3.1415926;int main( )
{float x,y;cin>>x>>y;float r,angle,radian;r=sqrt(x*x+y*y);//半径radian=atan(y/x);//弧度angle=radian*180/pai;//角度printf("%.1f %.1f",r,angle);return 0;
}
MT1053
#include<bits/stdc++.h> const double pai = 3.1415926;
using namespace std;
//angle 角度
//radian 弧度
//(r,angle)=>(x,y)int main( )
{double r,angle;//半径和角度cin>>r>>angle;double x,y,radian;radian=(angle/180.0)*pai;//弧度x=r*cos(radian);y=r*sin(radian);printf("%.2lf,%.2lf",x,y);return 0;
}
MT1054
#include<bits/stdc++.h> using namespace std;int main( )
{double C,R,H;cin>>R>>H;C=2*(sqrt(H*(2*R-H)));printf("%.2lf",C);return 0;
}
MT1055
#include<bits/stdc++.h> using namespace std;int main( )
{double C,H,R;cin>>C>>H;R=(C*C+4*H*H)/(8*H);printf("%.2lf",R);return 0;
}
MT1056
#include<bits/stdc++.h> using namespace std;int main( )
{double R,A,L;cin>>R>>A;L=0.01745*R*A;printf("%.2lf",L);return 0;
}
MT1057
#include<bits/stdc++.h> using namespace std;int main( )
{double R,A,L;cin>>R>>L;A=57.296*L/R;printf("%.2lf",A);return 0;
}
MT1058
#include<bits/stdc++.h> using namespace std;int main( )
{double C,R,H;cin>>R>>C;H=R-0.5*(sqrt(4*R*R-C*C));printf("%.2lf",H);return 0;
}
MT1059
#include<bits/stdc++.h> using namespace std;int main( )
{double C,R,H,L,F;cin>>R>>C>>H>>L;F=0.5*(R*L-C*(R-H));printf("%.3lf",F);return 0;
}
MT1060
#include<bits/stdc++.h> const double PI=3.1415926;
using namespace std;int main( )
{double R,H,S;cin>>R>>H;S=PI*R*(R+sqrt(R*R+H*H));printf("Surface area=%.2lf",S);return 0;
}
MT1061
#include<bits/stdc++.h> const double PI=3.1415926;
using namespace std;int main( )
{double R,H,V;cin>>R>>H;V=(1.0/3)*(PI*R*R)*H;//注意:必须是1.0/3,不能是简单的1/3printf("%lf",V);return 0;
}
MT1063
#include<bits/stdc++.h> using namespace std;int main( )
{double r;cin>>r;printf("%lf",r*r*r);return 0;
}
MT1064
#include<bits/stdc++.h> using namespace std;int main( )
{double r;cin>>r;printf("%lf",r*r*6);return 0;
}
MT1065
#include<bits/stdc++.h> using namespace std;int main( )
{double c,k,g;cin>>c>>k>>g;printf("%lf",2*(c*k+c*g+k*g));return 0;
}
MT1066
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;cout<<2*n;return 0;
}
MT1067
#include<bits/stdc++.h> using namespace std;int main( )
{//两点确定一条之间.所以是C(n,2)//也就是n*(n-1)/2int n;cin>>n;cout<<n*(n-1)/2;return 0;
}
MT1068
#include<bits/stdc++.h> using namespace std;int main( )
{int n,s;cin>>n;s=n*(n+1)/2+1;cout<<s;return 0;
}
MT1069
#include<bits/stdc++.h> using namespace std;int main( )
{//规律:/*1->22->43->8...n->n(n-1)+2*/int n;cin>>n;cout<<n*(n-1)+2;return 0;
}
MT1071
#include<bits/stdc++.h> using namespace std;int main( )
{printf("%lf",(45.0/2+43.41592)-36.0/7);return 0;
}
MT1073
#include<bits/stdc++.h> using namespace std;int main( )
{int N,M;cin>>N>>M;int answer;answer=(N+M)%12;printf("%3d",answer);return 0;
}
MT1074
#include<bits/stdc++.h> using namespace std;int main( )
{//ln是log()//lg是log10()double n;cin>>n;printf("%.2lf %.2lf %.2lf",sqrt(n),log(n),log10(n));return 0;
}
MT1075
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;//floor()是向下取整,//因为结果要求输出整型,//但是floor()的返回值是double类型,//所以结果要进行int强转printf("%d",int( floor(sqrt(n)) ));return 0;
}
MT1076
#include<bits/stdc++.h> using namespace std;int main( )
{double x1,y1,z1,x2,y2,z2;cin>>x1>>y1>>z1>>x2>>y2>>z2;double x,y,z;x=x1-x2;y=y1-y2;z=z1-z2;printf("%.2lf",sqrt(x*x+y*y+z*z));return 0;
}
MT1077
//做法一:
/*#include<bits/stdc++.h> using namespace std;int main( )
{int n;bool flag=true;cin>>n;while(n%10){int tmp = n%10;if(tmp!=4 && tmp!=7){flag=false;break;}n=n/10;}if(flag)cout<<"YES";else cout<<"NO";return 0;
}*///做法二:
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;while(n%10){int tmp = n%10;if(tmp!=4 && tmp!=7){cout<<"NO";return 0;}n=n/10;}cout<<"YES";return 0;
}
MT1079
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;for(int i=1;i<=n;i++){cout<<i<<" "<<floor(n/i)<<endl;}return 0;
}
MT1080
#include<bits/stdc++.h> using namespace std;int main( )
{int n;int sum = 0;cin>>n;for(int i=1;i<=n;i++){sum += i*floor(n/i);}cout<<sum;return 0;
}
MT1081
#include<bits/stdc++.h> using namespace std;int main( )
{int n;int sum;cin>>n;for(int i=1;i<=n;i++){sum += (i+floor(n/i));}cout<<sum;return 0;
}
MT1082
#include<bits/stdc++.h> using namespace std;int main( )
{int n;int sum;cin>>n;for(int i=1;i<=n;i++){sum += (i-floor(n/i));}cout<<sum;return 0;
}
MT1083
#include<bits/stdc++.h> using namespace std;int main( )
{/*还是用到pow函数,求立方根其实就是求1/3次方//特别注意:向下取整floor()函数的返回值是double,所以要写成1.0/3//另外,答案要整型,最后再int强制类型转换一下*/int n;cin>>n;printf("%3d", int( floor( pow(n,1.0/3) ) ) );return 0;
}
MT1085
#include<bits/stdc++.h> using namespace std;int main( )
{//数据给的是年利率,//让我们求月利率;//然后再求每月分期double p,r1,r2,n;//r1年利率,r2月利率cin>>p>>r1>>n;r2=r1/(12*100);double EMI;EMI=p*r2*pow(1+r2,n)/( pow(1+r2,n)-1 );printf("%.1lf",EMI);return 0;
}
MT1086
#include<bits/stdc++.h> using namespace std;int main( )
{double km;cin>>km;printf("%.2lf公里=%.2lf米",km,km*1000.0);return 0;
}
MT1087
#include<bits/stdc++.h> using namespace std;
/*
摄氏温度与华氏温度的转换:
1C=(1F-32)*1.8
1F=1C*1.8+32
*/
int main( )
{double c;cin>>c;printf("%.2lf C=%.2lf F", c, 1.0*c*1.8+32.0);return 0;
}
MT1088
#include<bits/stdc++.h> using namespace std;int main( )
{//c=9(f-32)/5;double f;cin>>f;printf("%lf",9.0*(f-32)/5);return 0;
}
MT1090
#include<bits/stdc++.h> using namespace std;int main( )
{//4n-2(n+1)-1//-1 1 3 5 7 ...int n;cin>>n;printf("%d",4*n-2*(n+1)-1);return 0;
}
MT1096
#include<bits/stdc++.h> using namespace std;int main( )
{//S(max)=r*r/4; //r是斜边//=> r=sqrt(4*S);int S;cin>>S;int r;//sqrt(4*S)得到小数,注意强制类型转换printf("%d",int( sqrt(4*S) ));return 0;
}
MT1097
#include<bits/stdc++.h> using namespace std;//n条直线可以把一个平面分成多少个部分
/*
n(n+1)/2+1
*/
int main( )
{int n;cin>>n;cout<<n*(n+1)/2+1;return 0;
}
MT1100
#include<bits/stdc++.h> #define f(x) (x*x)
using namespace std;int main( )
{printf("%lf",f(9.0)/f(6.0));return 0;
}
MT1101
#include<bits/stdc++.h> #define f(x) ( (x) *( (x) -1))
using namespace std;int main( )
{int a,b;cin>>a>>b;printf("%lf",f(a*1.0+b*1.0));return 0;
}
MT1102
#include<bits/stdc++.h> #define V(x,y,z) ((x)*(y)*(z))using namespace std;int main( )
{double a,b,c;cin>>a>>b>>c;printf("%lf",V(a,b,c));return 0;
}
MT1103
#include<bits/stdc++.h> #define V(r) ( (4.0/3)*(3.14159)*r*r*r )
using namespace std;int main( )
{double r;cin>>r;printf("%lf",V(r));return 0;
}
MT1105
#include<bits/stdc++.h> #define f(a,b,c) a*2.54,b*0.31,c*1.61using namespace std;int main( )
{double x,y,z;cin>>x >>y >>z;printf("%.2lf %.2lf %.2lf",f(x,y,z));return 0;
}
MT1107
#include<bits/stdc++.h> #define f(x) x*3.79
using namespace std;int main( )
{double x;cin>>x;printf("%.2lf",f(x));return 0;
}
MT1108
#include<bits/stdc++.h> using namespace std;int main( )
{double x;int n;//fixed可以消除科学计数法//fixed与setprecision(n)连用可以控制小数点后的位数,//现在就可以理解因为那是定点数记数法。/*setw()若想用setw(int n)呢,要加头文件 < iomanip > 其控制后面输出的长度,默认右对齐,输出内容长度不够用空格补齐*//*总结:fixed与setprecision(n)连用相当于printf输出的%x.ylf的ysetw()相当于printf输出的%x.ylf的x大多数的时候,printf非常好用,但是有时候,printf并不能满足我们的需求,就需要fixed与setprecision(n)的配合*/cin>>x>>n;cout<<fixed<<setprecision(n)<<x<< endl;return 0;
}
MT1109
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;if(n<10){cout<<n<<"小于10 "<<endl;}else if(n==10){cout<<n<<"等于10 "<<endl;}else{cout<<n<<"大于10 "<<endl;}return 0;
}
MT1110
#include<bits/stdc++.h> using namespace std;int main( )
{int x,y;cin>>x>>y;if(x<y){cout<<x;}else{cout<<y;}return 0;
}
MT1111
#include<bits/stdc++.h> using namespace std;int main( )
{int x,y;cin>>x>>y;if(x>y){cout<<x;}else{cout<<y;}return 0;
}
MT1112
#include<bits/stdc++.h> using namespace std;int main( )
{int x,y,z;cin>>x>>y>>z;if((x<y && y<z)||(z<y && y<x))cout<<y;if((y<x && x<z)||(z<x && x<y))cout<<x;if((x<z && z<y)||(y<z && z<x))cout<<z;return 0;
}
MT1114
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;if(n%2==0){cout<<"Y";}else{cout<<"N";}return 0;
}
MT1115
#include<bits/stdc++.h> using namespace std;int main( )
{int x,y;cin >> x >> y;if( (x<y) && (x%2==0) )cout<<"YES";else cout<<"NO";return 0;
}
MT1116
#include<bits/stdc++.h> using namespace std;int main( )
{int x;cin>>x;if( x>=10 && x<=99)cout<<"YES";else cout<<"NO";return 0;
}
MT1117
#include<bits/stdc++.h> using namespace std;int main( )
{//判断负数个数是否大于等于2//注意:可能有0int x,y,z;cin>>x>>y>>z;int tmp=0;if(x<0)tmp++;if(y<0)tmp++;if(z<0)tmp++;if(tmp>=2)cout<<"YES";else cout<<"NO";return 0;
}
MT1119
#include<bits/stdc++.h> using namespace std;int main( )
{char ch;cin>>ch;//cout<<int('B')<<endl;//A-65 a-97//65~96都是大写//>=97都是小写if('a'<= ch && ch <='z'){//如果是小写,就变大写ch = ch-32;//小写变大写printf("%c\n",ch);}else if('A'<= ch && ch <='Z'){ch = ch+32;//大写变小写printf("%c\n",ch);}else{//一开始我还疑惑这个else与没有必要写,//再仔细看题的最后一句://其他非法输入(非字母的输入)则原样输出printf("%c",ch);}return 0;
}
MT1121
#include<bits/stdc++.h> using namespace std;int main( )
{int n;char rank;cin>>n;n = n/10;switch(n){case 0:case 1:case 3:case 4:case 5: rank='F';break;case 6: rank='D';break;case 7: rank='C';break;case 8: rank='B';break;case 9:case 10: rank='A';break;default: rank='S';}cout<<rank<<endl;return 0;
}
MT1122
#include<bits/stdc++.h> using namespace std;int main( )
{int n;char rank;cin>>n;if(90<=n && n<=100)rank='A';else if(80<=n &&n<=89)rank='B';else if(70<=n &&n<=79)rank='C';else if(60<=n &&n<=69)rank='D';else rank='F';cout<<rank<<endl;return 0;
}
MT1123
#include<bits/stdc++.h> using namespace std;int main( )
{char ch;//a o e i u 是元音cin>>ch;switch(ch){case 'a':case 'o':case 'e':case 'i':case 'u':case 'A':case 'O':case 'E':case 'I':case 'U':cout<<"Y"<<endl;break;default:cout<<"N"<<endl;}return 0;
}
MT1124
#include<bits/stdc++.h> using namespace std;int main( )
{//1~9int n;cin>>n;if(n == 1)cout<<"I";else if(n == 2)cout<<"II";else if(n == 3)cout<<"III";else if(n == 4)cout<<"IV";else if(n == 5)cout<<"V";else if(n == 6)cout<<"VI";else if(n == 7)cout<<"VII";else if(n ==8)cout<<"VIII";else cout<<"IX";return 0;
}
MT1127
#include<bits/stdc++.h> using namespace std;int main( )
{//1900是鼠年string str[12]={"rat", "ox", "tiger", "rabbit","dragon", "snake", "horse", "sheep", "monkey", "rooster", "dog", "pig"};int year;cin>>year;cout<<str[(year-1900)%12];return 0;
}
MT1128
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;switch(n){case 1:cout<<"6";break;case 2:cout<<"5";break;case 3:cout<<"4";break;case 4:cout<<"3";break;case 5:cout<<"2";break;case 6:cout<<"1";break;default: cout<<"-1";}return 0;
}
MT1129
#include<bits/stdc++.h> using namespace std;int main( )
{//正反面一定是7cout<<7;return 0;
}
MT1130
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;switch(n){case 1:cout<<1-6;break;case 2:cout<<2-5;break;case 3:cout<<3-4;break;case 4:cout<<4-3;break;case 5:cout<<5-2;break;case 6:cout<<6-1;break;}return 0;
}
MT1131
#include<bits/stdc++.h> using namespace std;/*
算术运算符: + - * /
关系运算符: = > <
逻辑运算符: ! & | ^
*/
int main( )
{char a;cin>>a;//cout<<int('0')<<endl;//48//cout<<int('9')<<endl;//57if(a>=48 &&a<=57)cout<<"Number";else if(a>=65 && a<=90)cout<<"Capital letter";else if(a>=97 && a<=122)cout<<"Lowercase letter";else if(a=='+' ||a=='-'||a=='*'||a=='/')cout<<"Arithmetic operators";else if(a=='=' ||a=='>'||a=='v')cout<<"Relational operators";else if(a=='!' ||a=='&'||a=='|'||a=='^')cout<<"Logical operators";else cout<<"Other character";return 0;
}
MT1132
#include<bits/stdc++.h> using namespace std;void trans(int a){//零//壹贰叁肆伍陆柒捌玖拾if(a==1)cout<<"壹";else if(a==2)cout<<"贰";else if(a==3)cout<<"叁";else if(a==4)cout<<"肆";else if(a==5)cout<<"伍";else if(a==6)cout<<"陆";else if(a==7)cout<<"柒";else if(a==8)cout<<"捌";else if(a==9)cout<<"玖";}int main( )
{/*范围是0~99*/char a[3];cin>>a;if(strlen(a)==1 && a[0]=='0'){cout<<"零";//只有一位,且为0的情况,即输出是0的情况}trans(a[0]-'0');//第一位不是0的情况,相当于把十位输出(如果有两位),(如果只有一位,相当于把个位输出)if(strlen(a)==2 && a[0]!='0'){//既然是两位,那上一行输出的就是十位了,记下来我们就把个位输出就好了cout<<"拾";//为甚要加上这个条件?? && a[0]!='0'//因为可能出现05这样第一位是0的情况}trans(a[1]-'0');//输出个位cout<<"元整"<<endl;return 0;
}
MT1133
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;double cost = 0;if(n<=5){cost = 10;}else if(n<=10){cost = 10+(double)(n-5)*2;}else{ cost = 10 + 2*5 + (double)(n-10)*2.5;}printf("%.2lf",cost);return 0;
}
MT1134
#include<bits/stdc++.h> using namespace std;int main( )
{char a;double x,y;scanf("%lf%c%lf",&x,&a,&y);if(a == '+')printf("%lf",x+y);else if(a =='-')printf("%lf",x-y);else if(a =='*')printf("%lf",x*y);else if(a == '/')printf("%lf",x/y);return 0;
}
MT1138
#include<bits/stdc++.h> using namespace std;
//非闰年
int days[12]={31,28,31,30,31,30,31,31,30,31,30,31};
int main( )
{//年月日,是否为闰年的标志,总天数int y,m,d,flag = 0,total = 0;scanf("%d-%d-%d",&y,&m,&d);//只能把4除尽,或者可以把400除尽 , 就是闰年if( (y%4==0 && y%100 !=0) || (y%400==0))flag=1;for(int i=0;i<m;i++){//把之前几个月的天数加起来total += days[i-1];//因为下标从0开始,所以要减一//比如2013-6-17//i<6 就先把0 1 2 3 4 5的天数加起来}if(flag==1 && m>2){//如果是闰年,且月份大于2,说明少加了一天,//那就多加一天total = total + d + 1;}else{//否则就没有少加,把剩下的加回来就行total = total + d;}cout<<total<<endl;return 0;
}
MT1139
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;if(n%11==0 ||n%13==0)cout<<"YES";else cout<<"NO";return 0;
}
MT1140
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;//判断是不是7的倍数if(n%7==0)cout<<"YES";else cout<<"NO";return 0;
}
MT1141
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;//判断是不是13的倍数if(n%13==0)cout<<"YES";else cout<<"NO";return 0;
}
MT1142
#include<bits/stdc++.h> using namespace std;int main( )
{int n,m,k;cin>>n>>m>>k;int count;for(int i=n;i<=m;i++){if(i%k==0)count++;}cout<<count;return 0;
}
MT1144
#include<bits/stdc++.h> using namespace std;//这个题又叫"哈沙德数"
//就是说:问:18能不能被1+8整除? 18/(1+8)==0 ,所以"YES
int main( )
{int num;cin>>num;int sum=0;int tmp=num;//因为num会改变,我们需要一个值,来保存num最初始的值while(num){sum += num%10;num = num/10;}if(tmp %sum == 0 )cout<<"YES";else cout<<"NO";return 0;
}
MT1146
#include<bits/stdc++.h> using namespace std;int main( )
{int x;for(int i=1;i<100;i++){x=i;if(x%3==2 && x%5==3 && x%7==2){cout<<x;break;}}/*也可以直接:printf("%d",(2*70+3*21+2*15)%105);*/return 0;
}
MT1148
#include<bits/stdc++.h> using namespace std;int main( )
{int x;//题目说>10for(int i=11;i<200;i++){x=i;if(x%3==2 && x%5==3 && x%7==1){cout<<x;break;}}return 0;
}
MT1149
#include<bits/stdc++.h> using namespace std;int main( )
{int x;//题目说>10for(int i=11;i<500;i++){x=i;if(x%7==2 && x%11==4 && x%13==5){cout<<x;break;}}return 0;
}
MT1150
#include<bits/stdc++.h> using namespace std;int main( )
{int x;//战死四五百人,一共1500人,还剩下1000~1100,问战死多少人for(int i=1000;i<1100;i++){x=i;if(x%3==2 && x%5==4 && x%7==6){cout<<1500-x;break;}}return 0;
}
MT1151
#include<bits/stdc++.h> using namespace std;int main( )
{int x;//题目说>10for(int i=11; ;i++){x=i;if(x%3==2 && x%5==2 && x%7==2){cout<<x;break;}}return 0;
}
MT1152
#include<bits/stdc++.h> using namespace std;int main( )
{int x;//题目说>10for(int i=11;i<200;i++){x=i;if(x%3==3-1 && x%5==5-1 && x%7==7-1){cout<<x;break;}}return 0;
}
MT1153
#include<bits/stdc++.h> using namespace std;//6=>1 2 3int main( )
{int n;int sum=0;cin>>n;for(int i=1;i<n;i++){if(n%i==0)sum = sum+i;}cout<<sum;return 0;
}
MT1154
#include<bits/stdc++.h> using namespace std;//6=>1 2 3int main( )
{int n;int sum=0;cin>>n;for(int i=1;i<n;i++){if(n%i==0)sum = sum+i;}if(sum<n)cout<<"YES";else cout<<"NO";return 0;
}
MT1155
#include<bits/stdc++.h> using namespace std;
/*
单位矩阵
|1 0 0|
|0 1 0|
|0 0 1|
*/int main( )
{int a[3][3];for(int i=0; i<=2;i++){for(int j=0; j<=2;j++){cin>>a[i][j];}}//对角线上的元素都为1,非对角线上的元素都为0if(a[0][0]==1 &&a[1][1]==1 && a[2][2]==1&&a[0][1]==0 &&a[0][2]==0&&a[1][0]==0 &&a[1][2]==0&&a[2][0]==0 &&a[2][1]==0)cout<<"YES";else cout<<"NO";return 0;
}
MT1157
#include<bits/stdc++.h> using namespace std;
/*
单位矩阵
|1 0 0|
|0 1 0|
|0 0 1|
*/int main( )
{int a[4][4] ,b[4][4];for(int i=0; i<=3;i++){for(int j=0; j<=3;j++){cin>>a[i][j];}}for(int i=0; i<=3;i++){for(int j=0; j<=3;j++){cin>>b[i][j];}}int flag=1;for(int i=0; i<=3;i++){for(int j=0; j<=3;j++){if(a[i][j]!=b[i][j]){flag=0;break;}}}if(flag==1)cout<<"YES";else cout<<"NO";return 0;
}
MT1159
#include<bits/stdc++.h>
using namespace std;int main( )
{int n;scanf("%d",&n);int x = 0;int count=0;//记录来自{4 5 6}集合的个数for(int i=1 ; i<=n ;i++){scanf("%d",&x);if( x==4 || x==5 || x==6){++count;printf("%d ",x);//输出整型,空格分隔。这后边有个空格别忘了}}if(!count){//输入的这几个数没一个来自{4 5 6}printf("-1");}return 0;
}
MT1161
#include<bits/stdc++.h> using namespace std;
/*
字符串转数字:atoi()或者stoi()
//atoi()的参数是 const char* ,
//因此对于一个字符串str我们必须调用 c_str()的方法把这个string转换成 const char*类型的,
//而stoi()的参数是const string*,不需要转化为 const char*;
数字转字符串:to_string()
*/
int main( )
{int n;cin>>n;//把0变5string str=to_string(n);for(int i=0;i<str.size();i++){if(str[i]=='0')str[i]='5';}int answer;answer = stoi(str);cout<<answer;return 0;
}
MT1163
#include<bits/stdc++.h> using namespace std;/*
质数(prime number)又称素数,有无限个。
一个大于1的自然数,除了1和它本身外,不能被其他自然数整除
*/
int judge(int i){if(i==1)return 0; for(int j=2;j<i ; j++){//除了1和它本身外,不能被其他自然数整除if(i % j == 0){return 0;//不是质数}}return 1;//是质数
}int main( )
{//两数都为质数,且他们的差为2.就叫孪生质数int x,y;cin>>x>>y;if(judge(x)==1 && judge(y)==1 && y-x==2){cout<<"YES";}else{cout<<"NO";}return 0;
}
MT1167
#include<bits/stdc++.h> using namespace std;
//76 * 76 = 5776 =>76 与 5776后两位数组相同,就称"自守数"
int main( )
{int n,nn;//nn表示n的平方cin>>n;nn = n*n;while(n){if(nn%10 != n%10){cout<<"NO";return 0;}n = n / 10;nn = nn / 10;}cout<<"YES";return 0;
}
MT1168
#include<bits/stdc++.h> using namespace std;/*
又是公式:
1x2x3x...x m == n (m的最大值是 n/2 +1 ,为啥要+1,因为n/2要向下取整,我们把他加回来)
eg:
1x2 == 2 (2是2的1/1)
1x2x3 == 6 (3是6的1/2)
1x2x3x4 == 12 (4是12的1/3)
1x2x3x4x5 == 60 (5是60的1/12)
...
所以,m最大值是n/2 +1
*/
int main( )
{int n;cin>>n;int sum=1;//存放乘积for(int i=1 ; i<=(n/2)+1 ; i++ ){sum = sum * i;if(sum == n){cout<<"YES";return 0;}}cout<<"NO";return 0;
}
MT1170
#include<bits/stdc++.h> using namespace std;int main( )
{//首先是4位数int n;cin>>n;if(n<1000 || n>9999){cout<<"NO";return 0;}//其次,每一位的4次方的和加起来 == 他本身int tmp;tmp = n;int sum = 0;while(tmp){sum += pow(tmp%10 , 4);tmp = tmp /10;}if(sum == n){cout<<"YES";}else{cout<<"NO";}return 0;
}
MT1171
#include<bits/stdc++.h> using namespace std;int main()
{
//1234=>1+2+3+4=10=>1+0=1int n;cin>>n;int tmp;while(n>=10){//当n<10,说明就是一位数了tmp = 0;while(n){tmp = tmp + n%10;n=n/10;}n = tmp;}if(n==1)cout<<"YES";else cout<<"NO";return 0;
}
MT1172
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;//每一位数的阶乘累加int sum = 0;int original = n;//记录n原来的数int tmp;while(n){tmp = 1;//每次都需要从1开始for(int i = 1; i<=n%10 ; i++)tmp *= i; sum += tmp;n = n/10;}if(original==sum)cout<<"YES";else cout<<"NO";return 0;
}
MT1173
#include<bits/stdc++.h> using namespace std;
/*
这个题比较坑的一点:
比如说:112与221是否符合魔数的定义呢?
答:不符合,因为112中,1出现2次,2出现1次;
221中,1出现1次,2出现2次;所以,就需要记录次数
*/
int main( )
{int n;//一定是个正整数cin>>n;int new_num = n*2;//记录新数map<int,int>m;while(n){m[n%10]++;n = n/10;}while(new_num){auto ret=m.find(new_num%10);//寻找new_num%10if(ret!=m.end()){//找到了if(m[new_num%10]!=0){m[new_num%10]--;}else{//数量不匹配cout<<"NO"<<endl;return 0;}}else{//没找到cout<<"NO"<<endl;return 0;}new_num = new_num/10;}cout<<"YES"<<endl;return 0;
}
//方二:
/*#include<bits/stdc++.h>
using namespace std;int main(){int a,b;int comp[2][10]={0};//第一行与第二行进行对比cin>>a;b=a*2;while(a){comp[0][a%10] += 1;a = a/10;}while(b){comp[1][b%10] += 1;b = b/10;}//注:如果不是魔数,但是数字都出现了,只是数字个数不一样,a%10与b%10也是一一对应的for(int i =0;i<=9;i++){if(comp[0][i]!=comp[1][i]){cout<<"NO";return 0;}}cout<<"YES";return 0;
}*/
MT1174
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;//n是否等于a的b次方if(n==1){cout<<"YES";return 0;}for(int i = 2;i<=n;i++){for(int j=2;pow(i,j)<=n;j++){if(pow(i,j)==n){cout<<"YES";return 0;}}}cout<<"NO";return 0;
}
MT1176
#include<bits/stdc++.h> using namespace std;int main( )
{int x1,x2,y1,y2;cin>>x1>>y1>>x2>>y2;//距离=sqrt( (x1-x2)的平方 +(y1-y2)的平方 )int tmp =pow((x1-x2),2) +pow((y1-y2),2);int distance = sqrt(tmp);if(distance*distance==tmp){//刚好是个整数,直接输出cout<<distance;}else{//不是整数,int默认向下取整,答案要求向上取整,那就+1cout<< distance+1;}return 0;
}
MT1178
#include<bits/stdc++.h> using namespace std;int main( )
{//判断第3个点是否在(前两个点组成的)线上int x1,x2,a,y1,y2,b;/*这个题的坑:就是输入格式大有讲究!!!服了!!!*/scanf("(%d,%d) (%d,%d)\n",&x1,&y1,&x2,&y2);scanf("(%d,%d)",&a,&b);/*可以比斜率,也可以比较 xz+yz是否==zy,如果相等,则z在直线上*//*double ans = sqrt( pow((x1-x2),2.0)+pow((y1-y2),2.0) );double ans1 = sqrt( pow((x1-a),2.0)+pow((y1-b),2.0) ) ;double ans2 = sqrt( pow((a-x2),2.0)+pow((b-y2),2.0) ) ;*/double ans = sqrt( pow((x1-x2),2)+pow((y1-y2),2) );double ans1 = sqrt( pow((x1-a),2)+pow((y1-b),2) ) ;double ans2 = sqrt( pow((a-x2),2)+pow((b-y2),2) ) ;if(ans == ans1+ans2)cout<<"YES";else cout<<"NO";return 0;
}
MT1183
#include<bits/stdc++.h> using namespace std;
struct Point//点 , 坐标
{int x,y;
};
struct Rect//矩形
{Point lt,rb;//矩形的左上角和右下角
};int main( )
{Rect rect,rect2,tmp;scanf("%d %d %d %d\n", &rect.lt.x,&rect.lt.y, &rect.rb.x,&rect.rb.y);scanf("%d %d %d %d", &rect2.lt.x,&rect2.lt.y, &rect2.rb.x,&rect2.rb.y);//矩形2在矩形1里面//判断1的"两个点"都在在2里边//第一个点if((rect.lt.x<rect2.lt.x&&rect.rb.x>rect2.lt.x&& rect.rb.y < rect2.lt.y &&rect.lt.y>rect2.lt.y)//1和2比,左上角x小y大 //第二个点&&(rect.lt.x <rect2.rb.x&&rect2.rb.x<rect.rb.x &&rect.rb.y<rect2.rb.y&&rect2.lt.y) ){cout<<"YES";return 0;}//矩形1在矩形2里面tmp = rect2;rect2 = rect;rect = rect2;if((rect.lt.x<rect2.lt.x&&rect.rb.x>rect2.lt.x&& rect.rb.y < rect2.lt.y &&rect.lt.y>rect2.lt.y)&&(rect.lt.x <rect2.rb.x&&rect2.rb.x<rect.rb.x &&rect.rb.y<rect2.rb.y&&rect2.lt.y) ){cout<<"YES";return 0;}cout<<"NO";return 0;
}
MT1184
#include<bits/stdc++.h> using namespace std;
/*
解决问题的核心:
两个矩形中心点在x,y轴上的投影长度"差的绝对值"
分别小于两个矩形长、宽和的一半,则相交
*/bool inside(int x1, int y1,int x2,int y2,int xk,int yk){if((xk == x1 || xk==x2)&&(yk>=y2 && yk<=y1))return 1;if((yk == y1 || yk==y2)&&(xk>=x1 && xk<=x2))return 1;return 0;
}int main( )
{int x1,x2,x3,x4,y1,y2,y3,y4;scanf("%d %d %d %d\n", &x1,&y1,&x2,&y2);scanf("%d %d %d %d", &x3,&y3,&x4,&y4);int r1_CenterX, r1_CenterY;//第一矩形的中心int r2_CenterX, r2_CenterY;//第二矩形的中心int r1_Lenght, r1_height;//矩形一的长宽int r2_Lenght, r2_height;//矩形二的长宽//算出中心点坐标r1_CenterX = (x1 + x2) / 2;r1_CenterY = (y1 + y2) / 2;r2_CenterX = (x3 + x4) / 2;r2_CenterY = (y3 + y4) / 2;//算出两个矩形的长宽r1_Lenght = abs(x1 - x2);r1_height = abs(y1 - y2);r2_Lenght = abs(x3 - x4);r2_height = abs(y3 - y4);//在本题中,只有一个交点视为相切,不相交//但是如果是一条边或者部分边重合就算相交了//相切不相交if(x1==x4 && y1==y4 || x2==x3 && y2==y3|| x1==x4 && y2==y3 || x2==x3 && y1==y4){cout<<"NO";return 0;}//如果是一条边或者部分边重合就算相交了//=>在排除相切不相交的前提下,只要有一个点在对方矩形上,就满足这个情况if(inside(x1,y1,x2,y2,x3,y3)||inside(x1,y1,x2,y2,x4,y4)||inside(x3,y3,x4,y4,x1,y1)||inside(x3,y3,x4,y4,x2,y2)){cout<<"YES";return 0;}//完全包含居然不算相交//1包含2或2包含1if(x1<x3 && y1>y3 && x2>x4 && y2<y4||x1>x3 && y1<y3 && x2<x4 && y2>y4){cout<<"NO";return 0;}//正常情况下,光有下面这个判断条件就OK了,但是这个破题,完全包含居然不算相交,//所以上边要先判断,是否是完全包含,完全包含居然不算相交//广义上的相交:if (abs(r1_CenterX - r2_CenterX) < ((r1_Lenght + r2_Lenght) / 2) && abs(r1_CenterY - r2_CenterY) < ((r1_height + r2_height) / 2)) {printf("YES");}else {printf("NO");}return 0;
}
MT1185
#include<bits/stdc++.h> using namespace std;int main( )
{for(int i=0;i<8;i++){cout<<i<<" ";}return 0;
}
MT1186
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;for(int i=n;i>=1;i--){cout<<i<<" ";}return 0;
}
MT1187
#include<bits/stdc++.h> using namespace std;int main( )
{int sum=0;for(int i=1;i<=10;i++){sum += i;}cout<<sum;return 0;
}
MT1188
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;double sum=0.0;double x;for(int i =0;i<n;i++){cin>>x;sum += x;}printf("%.2lf",sum/n);return 0;
}
MT1189
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;double sum1,sum2;double x;for(int i =0;i<n;i++){cin>>x;if(x>0)sum1 += x;else sum2 += x;}printf("%lf %lf",sum1,sum2);return 0;
}
MT1190
#include<bits/stdc++.h> using namespace std;int main( )
{int x1,x2,x3,x4;double s[5];for(int i=0;i<5;i++){scanf("%d/%d %d/%d",&x1,&x2,&x3,&x4);s[i] = (1.0*x1*x3)/(1.0*x2*x4);}for(int i=0;i<5;i++){printf("%lf\n",s[i]);}return 0;
}
MT1191
#include<bits/stdc++.h> using namespace std;int main( )
{ int n,m;cin>>n>>m;for(int i =0;i<m;i++){n *= 0.5; }cout<<n;return 0;
}
MT1192
#include<bits/stdc++.h> using namespace std;int main( )
{int n,m;cin>>n>>m;for(int i=0;i<m;i++){n *= 2;}cout<<n;return 0;
}
MT1193
#include<bits/stdc++.h> using namespace std;
//pow(2,2) pow(4,2) pow(8,2)
int main( )
{int n;int sum=0;cin>>n;for(int i=1;i<=n;i++){sum += pow(i*2,2);}cout<<sum;return 0;
}
MT1194
#include<bits/stdc++.h> using namespace std;
//35=1+9+25
int main( )
{int n;int sum=0;cin>>n;for(int i=1;i<=n;i++){sum += pow(2*i-1,2);}cout<<sum;return 0;
}
MT1195
#include<bits/stdc++.h>
using namespace std;int main( )
{int n,m;cin>>n>>m;double ans=0;for(int i=n;i<=m;i++){//ans += 1.0/(i*i);//分母一定要加括号ans += 1.0/pow(i,2);//cout<<ans<<endl;}printf("%.5lf",ans);return 0;
}
MT1196
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;int sum=1;for(int i=1;i<=n;i++){sum *= i;}cout<<n<<"!="<<sum;return 0;
}
MT1197
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;int sum=0;int tmp;for(int i=n;i>=1;i--){tmp=1;for(int j=1;j<=i;j++){tmp *= j;}sum += tmp;}cout<<sum;return 0;
}
MT1198
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;int sum=0;int tmp;for(int i=n;i>=1;i--){tmp=1;for(int j=1;j<=i;j++){tmp *= j;}sum -= tmp;//算出来的是-1!-2!-3!-...-n!}sum = sum+1+1;cout<<sum;return 0;
}
MT1199
#include<bits/stdc++.h> using namespace std;int main( )
{double n,r;cin>>n>>r;double tmp1=1,tmp2=1;for(double i = 1;i<=n;i++){tmp1 *= i;}for(double i = 1;i<=n-r;i++){tmp2 *= i;}//cout<< tmp1<<" "<<tmp2<<endl;printf("%.2lf",(tmp1)/(tmp2));return 0;
}
MT1201
#include<bits/stdc++.h> using namespace std;
//145 == 1!+4!+5!强数
int fun(int n){int sum =1;for(int i=1;i<=n;i++){sum *=i;}return sum;
}
int main( )
{int n;cin>>n;int tmp=n;int sum=0;while(n){sum += fun(n%10);n /= 10;}if(sum==tmp){cout<<"YES";}else{cout<<"NO";}return 0;
}
MT1203
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;for(int i=0;i<n;i++){for(int i=0;i<n;i++){printf("F ");}printf("\n");}return 0;
}
MT1204
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;for(int i=0;i<n;i++){for(int j=0;j<=2*i;j++){printf("F");}printf("\n");}return 0;
}
MT1206
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;for(int i=0;i<n;i++){for(int j=0;j<n-i-1;j++){printf(" ");}for(int j=0;j<=2*i;j++){printf("W");}printf("\n");}return 0;
}
MT1207
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;for(int i=0;i<n;i++){for(int j=0;j<i;j++){printf(" ");//0 1 2}for(int k=n-i;k>0;k--){printf("W ");}printf("\n");}return 0;
}
MT1208
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;for(int i=0;i<n;i++){for(int j=0;j<n-i-1;j++){printf(" ");//0 1 2}for(int k=0;k<=i;k++){printf("W ");}printf("\n");}for(int i=0;i<n-1;i++){for(int j=0;j<=i;j++){printf(" ");//0 1 2}for(int k=n-i-1;k>0;k--){printf("W ");}printf("\n");}return 0;
}
MT1211
#include<bits/stdc++.h>
using namespace std;
int main( )
{
int n;
cin>>n;
for(int i=1;i<=9;i++){
cout<<n*i<<" ";
}
return 0;
}
MT1212
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;for(int i=1;i<=n;i++){for(int j=1;j<=i;j++){printf("%d*%d=%-2d ",i,j,j*i);//%-2d留两个空间}cout<<endl;}return 0;
}
MT1213
#include<bits/stdc++.h> using namespace std;
//3 1
//1+11+111==123
int main( )
{int n,num;cin>>n>>num;int sum = 0;int tmp = num;for(int i=1;i<=n;i++){sum = sum + tmp;tmp = tmp*10+num;//cout<<sum<<" "<<tmp<<endl;}cout<<sum;return 0;
}
MT1214
#include<bits/stdc++.h> using namespace std;int main( )
{//本题就是让你输出一个上三角这样的格式int a[10];for(int i = 1;i<=9;i++){cin>>a[i];}for(int i = 1 ; i<=9 ; i++ ){if(i==4 || i==7 || i==8)cout<<0<<" ";else cout<<a[i]<<" ";if(i==3 || i==6)cout<<endl;}return 0;
}
MT1215
#include<bits/stdc++.h> using namespace std;int main( )
{//本题就是让你输出一个下三角这样的格式int a[10];for(int i = 1;i<=9;i++){cin>>a[i];}for(int i = 1 ; i<=9 ; i++ ){if(i==2 || i==3 || i==6)cout<<0<<" ";else cout<<a[i]<<" ";if(i==3 || i==6)cout<<endl;}return 0;
}
MT1216
#include<bits/stdc++.h> using namespace std;int main( )
{int n;int a[100];int b[100];cin>>n;for(int i=0; i<n; i++)cin>>a[i];for(int i=0; i<n; i++)cin>>b[i];for(int i=0; i<n; i++)cout<<a[i]+b[i]<<" ";return 0;
}
MT1217
#include<bits/stdc++.h> using namespace std;int main( )
{int A[3][4]={0},B[4][3]={0},C[3][3]={0};for(int a=0 ; a<3; a++){for(int b=0; b<4; b++){cin>>A[a][b];}}for(int a=0 ; a<4; a++){for(int b=0; b<3; b++){cin>>B[a][b];}}for(int i=0 ; i<3; i++){for(int j=0; j<3; j++){for(int m=0;m<4;m++){C[i][j] += A[i][m] * B[m][j];}}}for(int a=0;a<3;a++){for(int b = 0; b<3;b++){printf("%3d ",C[a][b]);}cout<<endl;}return 0;
}
MT1219
#include<bits/stdc++.h> using namespace std;int main( )
{//1 1 2 3 5 8 ...int f1=1,f2=1,fn=1;int n;cin>>n;for(int i =1 ;fn<=n//判断条件就是,只要fn小于等于n,就一直循环;){if(fn==n){cout<<"YES";return 0;}fn = f1+f2;f1 = f2;f2 = fn;}cout<<"NO";return 0;
}
MT1220
/*#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;int f1=1,f2=1,fn;if(n==1){cout<<1;return 0;}if(n==2){cout<<1<<" "<<1;return 0;}cout<<1<<" "<<1<<" ";for(int i=3;i<=n;i++){// 1 1 2 3 5 8 ...fn=f1+f2;f1=f2;f2=fn;//cout<<endl;//cout<<f1<<" "<<f2<<endl;cout<<fn<<" ";}return 0;
}*/
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;int f[50]={0};f[1]=f[2]=1;if(n<=2){for(int i=1;i<=n;i++){cout<<f[i]<<" ";}return 0;}//n>=3for(int i = 3;i<=n;i++){f[i]=f[i-1]+f[i-2];}for(int i =1;i<=n;i++){cout<<f[i]<<" ";}return 0;
}
MT1222
#include<bits/stdc++.h> using namespace std;int main( )
{int n,m,k;//项数,首项,差值cin>>n>>m>>k;int sum = 0;int tmp;for(int i=0;i<n;i++){tmp = m + k*i;sum += tmp;}cout<<sum;return 0;
}
MT1223
#include<bits/stdc++.h> using namespace std;// 3 10 21 36 ...
// 7 11 15 =>差为4的等差数列
int main( )
{int A[10000],k=7,n;//n小于1万,k是差值cin>>n;A[1]=3;for(int i=2;i<=n;i++){//先存A[i] = A[i-1]+k;k += 4;}cout<<A[n];//再取return 0;
}
MT1224
#include<bits/stdc++.h> using namespace std;
//规律:找正方形,长方形不算
/*
1x1 => 1
2x2 => 4+1=5
3x3 => 9+4+1
4x4 => 16+9+4+1...nxn => n方+(n-1)方+...+1方
*/
int main( )
{//备注提示我们:用longlong n;cin>>n;long sum=0;for(int i=1;i<=n;i++){sum += pow(i,2);}cout<<sum;return 0;
}
MT1226
#include<bits/stdc++.h> using namespace std;int main( )
{int a,b;//左右象限cin>>a>>b;double denominator=0,sum=0;//分母和式子总和int l=0,r=0;//n范围的左右象限for(int i=1;i<=100000;i++){//备注提醒了,n最大不超过10万//分母的规律:1 1+1/2 1+1/2+1/3denominator += 1.0/i;sum += 1.0/denominator;if(sum>=b)break;//我们要的是超越之前的数据if(sum>a){if(l==0)l=i;r=i;}}cout<<l<<" "<<r;return 0;
}
MT1228
#include<bits/stdc++.h> using namespace std;
//爬楼梯,动态规划
int main( )
{int n;cin>>n;/*抢0个糖果有0种,抢1个糖果有1种,抢2个糖果有2种*/int a[10]={0,1,2};for(int i=3;i<=n;i++){a[i] = a[i-1]+a[i-2];}cout<<a[n];return 0;
}
MT1230
#include<bits/stdc++.h> using namespace std;int main( )
{double x=0,pi=0;for(int i=1;;i +=2 ){x = 1/(double)i;if(x<1e-6)break;if( (i+1)%4 == 0){//3,7,11...加1为4的倍数,是负数x=-x;}pi += x;}printf("%.2lf",4*pi);return 0;
}
MT1232
#include<bits/stdc++.h> using namespace std;int main( )
{int n ;cin>>n ;int sum=0;while(n){sum += n%10;n /= 10;}cout<<sum;return 0;
}
MT1233
#include<bits/stdc++.h> using namespace std;int main( )
{int n ;cin>>n ;int sum=0;while(n){sum += n%10;n /= 10;}cout<<sum;return 0;
}
MT1234
#include<bits/stdc++.h> using namespace std;int main( )
{ int n ;cin>>n ;int sum;while(n>=10){sum = 0;while(n){sum += n%10;n /= 10;}n=sum;}printf("%3d",n);return 0;
}
MT1236
#include<bits/stdc++.h> using namespace std;int main( )
{//判断是否为回文//用旧数组成新数,看新旧两数是否相等long long n;//看备注,长整型cin>>n;long long sum = 0;while(n){sum += n%10;n /= 10;}n=sum;//拿到各位数字的和,就判断他是不是回文long long temp,newa=0;temp=n;//将n复制一份,备份while(temp){newa = newa*10 + temp%10;temp = temp/10;}if(newa == n)cout<<"YES";else cout<<"NO";return 0;
}
MT1239
#include<bits/stdc++.h> using namespace std;
//407 == pow(4,3)+pow(0,3)+pow(7,3)
int main( )
{int n;cin>>n;int sum = 0;int temp = n;//备份原数while(n){sum += pow(n%10,3);n = n/10;}if(sum == temp)cout<<"Y";else cout<<"N";return 0;
}
MT1241
#include<bits/stdc++.h> using namespace std;int main( )
{string a;cin>>a;//string类的.size()与.length()都是返回长度for(int i=9; i>=0; i--){//把可能是9的遍历一遍,然后把可能是8的遍历一遍,...for(int j=0;j<a.size();j++){if( (a[j] -'0') == i)cout<<a[j];}}return 0;
}
MT1242
#include<bits/stdc++.h> using namespace std;int main( )
{//n分解成3个数//把前两个数的范围确定,第三个数的范围也就确定了int n;cin>>n;int count=0;//eg:5/*先把i为5的遍历完在把i为4的遍历完在把i为3的遍历完在把i为2的遍历完在把i为1的遍历完*///方一:/*for(int i = 0 ; i <= n ; i++){for(int j = i; j <=n-i ; j++){for(int k = j; k <=n-i-j ; k++){if(i+j+k == n){//cout<<i<<" "<<j<<" "<<k<<endl;count++;}}}}*///方二://i<<j<<k//k=n-i-j//所以,n-i-j>>j>>i//j>>i已经满足//只需要再满足n-i-j>>j即可for(int i = 0 ; i <= n ; i++){for(int j = i; j <=n-i ; j++){if( n-i-j >= j ){count++;}}}cout<<count;return 0;
}
MT1244
#include<bits/stdc++.h> using namespace std;
//(5,13)->(5,3)走了13/5=2步,13变成了13%5=3
int main( )
{int a,b;cin >> a >> b;int ans = 0;while(a !=0 && b!=0){//a,b有一个出现0就说明循环结束if(a<b)swap(a,b);//我们始终让a是最大的那个数ans = ans + a/b;a = a%b;}cout<<ans;return 0;
}
MT1246
#include<bits/stdc++.h> using namespace std;int main( )
{//x年初 y是年末 n是年利率=月利率*12//x*(1+n) -1000 =y //x == (y+1000)/(1+n) 年初的钱==(年末的钱+1000)/(1+n)double n=1+0.0063*12;//这里0.0063*12是年利率double money = 0;for(int a=1;a<=5;a++){money += 1000;money /= n;}cout<<money;return 0;
}
MT1247
#include<bits/stdc++.h> using namespace std;int main( )
{//x年初 y是年末 n是年利率=月利率*12//x*(1+n) -1000 =y //x == (y+1000)/(1+n) 年初的钱==(年末的钱+1000)/(1+n)double n=1+0.0058*12;//这里0.0058*12是年利率double money = 0;for(int a=1;a<=10;a++){money += 1000;money /= n;}cout<<money;return 0;
}
MT1249
#include<bits/stdc++.h> using namespace std;int main( )
{//输入3个非负整数a,b,c表示每种队形队尾的人数(a<3,b<5,c<7)int a,b,c;cin>>a>>b>>c;for(int i=10;i<=100;i++){//10~100if(i%3==a && i%5==b && i%7==c){cout<<i;return 0;}}cout<<-1;return 0;
}
MT1250
#include<bits/stdc++.h> using namespace std;int main( )
{int n;//台阶不超过1000阶。for(int i=1;i<=1000;i++){if(i%2==1 && i%3==2 && i%5==4 && i%6==5 &&i%7==0){cout<<i;return 0;}}return 0;
}
MT1252
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin>>n;int tmp=n;while(n){//(n%10)==0除数不能为0!if((n%10)==0 || tmp%(n%10)!=0 ){cout<<"NO";return 0;}n=n/10;}cout<<"YES";return 0;
}
MT1254
#include<bits/stdc++.h> using namespace std;int main( )
{int a,b;cin >> a >> b;for(int i=b;i>=a;i--){//从大到小bool flag=true;int tmp=i;int n=i;while(n){//(n%10)==0 除数不能为0!if((n%10)==0 || tmp%(n%10)!=0 ){//不是自除数flag=false;break;//跳出while循环}n=n/10;}if(flag==true){//是自除数cout<<tmp;return 0;}}return 0;
}
MT1255
#include<bits/stdc++.h> using namespace std;int main( )
{int a,b;cin >> a >> b;for(int i=a;i<=b;i++){//从大到小bool flag=true;int tmp=i;int n=i;while(n){//(n%10)==0 除数不能为0!if((n%10)==0 || tmp%(n%10)!=0 ){//不是自除数flag=false;break;//跳出while循环}n=n/10;}if(flag==true){//是自除数cout<<tmp;return 0;}}return 0;
}
MT1256
#include<bits/stdc++.h> using namespace std;
/*
1. n,m
2. k%m==0 k>n
3. %3d
*/
int main( )
{int n,m;cin>>n>>m;for(int k = n+1;k<10000;k++){//k要大于n,则k最小为n+1if(k%m==0){printf("%3d",k);return 0;}}return 0;
}
MT1257
#include<bits/stdc++.h> using namespace std;int main( )
{int ge_wei = 0;int count = 0;for(int i=10005;i<=99995;i++){ge_wei=i%10;if(ge_wei==5 && i%3==0)count++;}cout<<count;return 0;
}
MT1260
#include<bits/stdc++.h> using namespace std;int main( )
{int x,y,a,b,L;//x,y是坐标;a,b是跳的次数,总长为Lcin>>x>>y>>a>>b>>L;/*不可能遇不到,只是时间长短问题,除非速度一样*/if(a==b){cout<<-1;return 0;}int time=0;//(x+time*a)%L == (y+time*b)%Lwhile(1){if((x+time*a)%L == (y+time*b)%L){cout<<time;return 0;}time++;}return 0;
}
MT1262
#include<bits/stdc++.h> using namespace std;int main( )
{for(int i=999;i>=100;i--){if(555555%i==0){cout<<i;return 0;}}return 0;
}
MT1263
#include<bits/stdc++.h> using namespace std;int main( )
{int n;cin >> n;for(int i=999;i>=100;i--){if(n%i==0){cout<<i;break;}}return 0;
}
MT1264
#include<bits/stdc++.h> using namespace std;int main( )
{for(int i=1;i<=1000;i++){int sum=0;for(int j=i-1;j>=1;j--){//不包括自己j=i-1if(i%j==0)sum+=j;}if(sum==i)cout<<i<<" ";}return 0;
}
MT1265
#include<bits/stdc++.h> using namespace std;int main( )
{int a,b;cin>>a>>b;for(int i=a;i<=b;i++){int sum=0;for(int j=i-1;j>=1;j--){//不包括自己j=i-1if(i%j==0)sum+=j;}if(sum==i)cout<<i<<" ";}return 0;
}
MT1266
#include<bits/stdc++.h> using namespace std;int main( )
{deque<string>q;for(int i=1;i<=1000;i++){int sum=0;for(int j=i-1;j>=1;j--){//不包括自己j=i-1if(i%j==0)sum+=j;}if(sum==i){cout<<i<<"=";for(int j=1;j<=i-1;j++){//不包括自己j=i-1if(i%j==0){q.push_back(to_string(j));q.push_back("+");}}q.pop_back();//把多出来的加号pop出while(!q.empty()){cout<<q.front();q.pop_front();}cout<<endl;}}return 0;
}
MT1267
#include<bits/stdc++.h> using namespace std;
//12的因子有1 2 3 4 6 12 => 16+12>24 不是亏数
int main( )
{int n;cin>>n;int sum = 0;for(int i=1;i<=n;i++){if(n%i==0)sum+=i;}if(sum<2*n)cout<<"YES";else cout<<"NO";return 0;
}
MT1268
#include<bits/stdc++.h> using namespace std;
//12的因子有1 2 3 4 6 12
int main( )
{int n;cin>>n;vector<int>v;for(int i=1;i<=n;i++){if(n%i==0)v.push_back(i);}int sum = 0;vector<int>::iterator it;for(it=v.begin();it!=v.end();it++){//遍历因数for(int i=1;i<=(*it);i++){if((*it)%i==0){//cout<<i<<" ";sum+=i;}}}//cout<<endl;cout<<sum;return 0;
}
MT1269
#include<bits/stdc++.h> using namespace std;int main( )
{//素数:除了1和它本身外,没有其他公约数int a,b;cin>>a>>b;for(int i = a ; i <= b ; i++){if(i==1)continue;//1不是素数!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!bool flag = true;for(int j = 2 ; j <= i-1 ; j++){if(i%j==0){//不是素数flag = false;break;}}if(flag==true)cout<<i<<" ";}return 0;
}
MT1270
#include<bits/stdc++.h> using namespace std;int main( )
{//素数:除了1和它本身外,没有其他公约数int a,b;cin>>a>>b;int sum = 0;int k=1;for(int i = b ; i >= a ; i--){//注意看题:什么叫降序排列???就是从大到小if(i==1)continue;//1不是素数!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!bool flag = true;for(int j = 2 ; j <= i-1 ; j++){if(i%j==0){//不是素数flag = false;break;}}if(flag==true){//cout<<i<<" ";if((k%2)==1){//奇数加,偶数减sum = sum + i;}else{sum = sum - i;}//cout<<sum<<" ";k++;}}//cout<<endl;cout<<sum;return 0;
}
MT1272
#include<bits/stdc++.h> using namespace std;
/*
1不是双素数
两个素数的乘积所得的自然数是双素数
*/
bool isPrime(int n){if(n==1){return false;}for(int i=2;i<=sqrt(n);i++){if(n%i==0){return false;}}return true;
}
int main( )
{int a;cin>>a;if(isPrime(a)){cout<<"NO";}else{for(int i = 2 ;i<a;i++){if(a%i==0){int s = a / i;//判断两个乘数是不是素数if(isPrime(i)&&isPrime(s)){cout<<"YES";break;}else{cout<<"NO";break;}}}}return 0;
}
MT1273
#include<bits/stdc++.h> using namespace std;
/*
1不是双素数
两个素数的乘积所得的自然数是双素数
*/
bool isPrime(int n){if(n==1){return false;}for(int i=2;i<=sqrt(n);i++){if(n%i==0){return false;}}return true;
}
int main( )
{int n;cin>>n;for(int a=4;a<=n;a++){if(isPrime(n)){continue;}else{for(int i = 2 ;i<a;i++){if(a%i==0){int s = a / i;//判断两个乘数是不是素数if(isPrime(i)&&isPrime(s)){cout<<a<<" ";break;}/*else{cout<<"NO";break;}*/}}}}return 0;
}
MT1274
#include<bits/stdc++.h> using namespace std;int main( )
{//蓝桥杯直接从小到大手算//2 3 5 7 11 13 17 19 23 29//5 11 17 23 26cout<<"5 11 17 23 29"<<endl;return 0;
}
MT1275
#include<bits/stdc++.h> using namespace std;int main( )
{//蓝桥杯直接从小到大手算//2 3 5 7 11 13 17 19 23 29//3 13 23cout<<"3 13 23"<<endl;return 0;
}
MT1276
#include<bits/stdc++.h> using namespace std;
bool prime(int n){//模板记下来if(n==1){return false;}for(int i=2;i<=sqrt(n);i++){if(n%i==0){return false;}}return true;
}
int main( )
{int n;//n>2cin>>n;//仅输出最小值所在的组合,你品你细品for(int i=2;i<=n/2;i++){if(prime(i)&&prime(n-i)){cout<<i<<" "<<n-i;return 0;}}return 0;
}
MT1277
#include<bits/stdc++.h> using namespace std;
bool prime(int n)
{if(n==1)return false;for(int i = 2;i<=sqrt(n);i++){if(n%i==0){return false;}}return true;
}
int main( )
{//cout<<1671800;//一个世纪的100个年号是0到99for(int i = 0; ;i += 100){bool flag = true;for(int j=i ; j<i+100 ; j++){if(prime(j)){//是素数flag=false;break;}}if(flag==true){//j从[j+0,j+100]遍历完,都不是素数,则为答案cout<<i<<endl;break;}}return 0;
}
MT1278
#include <bits/stdc++.h>using namespace std;int main()
{int n;cin >> n;int tmp = n * n * n;//找到第一个奇数,+2即为下一个奇数for (int i = 1; i < tmp; i++){if ((i % 2 != 0)){/*n=33*3*3==27==7+9+11==7*3 +0*2 + 1*2 + 2*2n=22*2==4==1+3==1*2 + 0*2 +1*2n=xx*x*...*x==m*n + 0*2 + 1*2 + 2*2 +...+(n-1)*2 == m*n +((0+n-1)*n/2)*2*/if ( i * n + ((n - 1) * n / 2)*2 == tmp){ //连续n个奇数之和为tmp//是我们要找的那个数cout << n << "*" << n << "*" << n << "=";/*<<i<<"+"<<(i+2)<<"+"<<(i+4)*/deque<string>q;for(int j = 0;j<=n-1;j++){int tmp = i + j*2;q.push_back(to_string(tmp));q.push_back("+");}q.pop_back();while(!q.empty()){cout<<q.front();q.pop_front();}cout<<"="<<tmp;return 0;}}}return 0;
}
MT1280
#include<bits/stdc++.h> using namespace std;int main( )
{int x,y;cin>>x>>y;int tmp = 1;//每乘一个,就取一下后两位(本题与后两位以上没有关系)for(int i =1;i<= y;i++){tmp = tmp*x;tmp = tmp%100;}cout<<tmp;return 0;
}
MT1282
#include<bits/stdc++.h> using namespace std;
/*
这又是一道理解阅读题:
89=8的1次方+9的2次方
135 = 1的一次方 + 3的二次方 + 5的三次方
*/
int main( )
{int n;cin>>n;int tmp = n;int tmp2 = n;int sum = 0;int wei_shu = 0;while(n){//n用来算位数wei_shu ++; n /= 10;}//从后往前while(tmp2){//tmp2用来算位数次幂之和sumsum += pow(tmp2%10,wei_shu--);tmp2 = tmp2 / 10; }if(tmp == sum){//tmp用来和sum对比cout<<"YES";}else{cout<<"NO";}return 0;
}
MT1285
#include<stdio.h>int main()
{//对7取余,如果余0或余1,则可能是忠实数,//具体是不是还有进行下一步的判断。//但是,如果余1或0,以外的数,则一定不是忠实数int n;scanf("%d",&n);while(n/7) { //一直循环,直到剥壳,剥出最后的1~6,//(注意看,这里是没机会剥出0的),//判断条件之所以是while(n/7),而不是while(n),//是为了保留最后一次的余数,进行判断//次幂的特点就是:取整为0的前一次,一定是1,(先变1再变0)if(n%7==0) {//printf("->%d\n",n%7);n/=7;//printf("%d\n",n);} else if(n%7==1) {n/=7;//n=(n-1)/7;} else {printf("NO\n",n);return 0;}}if(n==1)//次幂就是有这样的特点,最后都会剩1//100/10=》10,10,1,0//49/7=》7,1,0printf("YES\n",n);elseprintf("NO\n",n);return 0;
}
MT1286
#include<stdio.h>int isZSS (int n)
{//对7取余,如果余0或余1,则可能是忠实数,//具体是不是还有进行下一步的判断。//但是,如果余1或0,以外的数,则一定不是忠实数while(n/7) { //一直循环,直到剥壳,剥出最后的1~6,//(注意看,这里是没机会剥出0的),//判断条件之所以是while(n/7),而不是while(n),//是为了保留最后一次的余数,进行判断//次幂的特点就是:取整为0的前一次,一定是1,(先变1再变0)if(n%7==0) {//printf("->%d\n",n%7);n/=7;//printf("%d\n",n);} else if(n%7==1) {n/=7;//n=(n-1)/7;} else {return 0;}}if(n==1)//次幂就是有这样的特点,最后都会剩1//100/10=》10,10,1,0//49/7=》7,1,0return 1;elsereturn 0;
}
int main()
{int n;//printf("请输入:\n");scanf("%d",&n);//printf("输入完毕\n");int cnt=0;for(int i=1; ;i++){if(isZSS(i)){if(++cnt==n){//输出第n个忠实数printf("%d", i);break;}}}return 0;
}
MT1287
#include<bits/stdc++.h> using namespace std;
int a[5];
int main( )
{for(int i=0;i<5;i++){scanf("%d",&a[i]);}for(int i=0;i<5;i++){if(i%2==0){a[i] += 97;}else{a[i] += 65;}}for(int i=0;i<5;i++){printf("%c ",a[i]);}return 0;
}
MT1289
#include<bits/stdc++.h> using namespace std;int main( )
{//cout<<33617;double sum = 0;for(double i = 1;;i++){sum += 1/i;if(sum>11 && sum<12){cout<<i;return 0;}}return 0;
}
MT1290
#include<bits/stdc++.h> using namespace std;int main( )
{int n;//1~ncin>>n;double sum = 0;for(int j=1;j<=n;j++){sum += (pow((-1),j-1)*pow(2,j))/( ( pow(2,j)+pow((-1),j) )* ( pow(2,(j+1))+pow(-1,(j+1)) ) );}printf("%.6lf",sum);return 0;
}
MT1291
#include<bits/stdc++.h> using namespace std;int jk(int i){if(i%2==1) return -1;else return 1;
}
int main( )
{int n;cin>>n;int pow=1,po=1;double sum=0;for(int j=1;j<=n;j++){pow*=2;po*=3;sum+=((double)(jk(j-1))*pow/2*po)/((double)( po+pow*(jk(j)) )*(3*po+(jk(j+1))*pow*2));}printf("%lf",sum);return 0;
}
MT1294
#include<bits/stdc++.h> using namespace std;int main( )
{char str[100];cin.getline(str, 100);int n;cin>>n;//strlen(str)不包括'\0'for(int i=n;i<=strlen(str)-1;i++)//减1是因为:数组从0开始,最大下标==长度-1cout << str[i];return 0;
}
MT1295
#include<bits/stdc++.h> using namespace std;int main( )
{//样例1是在下标6的前边进行插入char s1[100] , s2[100];int num;cin.getline(s1,sizeof(s1));scanf("%d\n",&num);cin.getline(s2,sizeof(s2));int n1=strlen(s1);int n2=strlen(s2);if(num >= n1){//则从s1末尾处接上s2字符串for(int i=n1;i<n2+n1;i++){//数组下标从0开始s1[i]=s2[i-n1];//一步一步往后移动}}else{//实现像样例1那样的效果for(int i=n1-1; i>=num ;i--){//先把s1的部分字符往后移动,给s2腾空间//从后往前一步一步的移动(把s1下标范围为[num,n1-1]的部分,从后往前依次移动n2步)//每次移动n2步s1[i+n2] = s1[i];}for(int i=0;i<n2;i++){//请君入瓮,将s2放入s1的指定位置s1[i+num]=s2[i];}}s1[n1+n2]='\0';cout<<s1;return 0;
}
MT1298
#include<bits/stdc++.h> using namespace std;int main( )
{int n,b[100],count=0;cin>>n;while(n){b[count]=n%2;//下标从小到大输入count++;n /= 2;}while(count){count--;//下标从大到小输出printf("%d",b[count]);}return 0;
}
MT1299
#include<bits/stdc++.h> using namespace std;
/*
101 1001
01234567 下标
12345678 个数1234 567 =>count0123 456 =>下标
->进入数组: 1001 101=>下标为4(count为5) 的字符输出时,应该加一个"空格"
->出数组: 101 1001
*/
int main( )
{for(int i =1;i<=5;i++){int n,b[100],count=0;scanf("%d",&n);while(n){b[count]=n%2;count++;n /= 2;}while(count){if((count-1)%4==0){count--;//个数减一变成下标cout<<b[count]<<" ";}else{count--;cout<<b[count];}}cout<<"\n";}return 0;
}
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