lc311309010103的专栏

#include int main() {     int t,n,m,a[1010],b[1010],i,j,c;     double sum=0.0;    while(scanf("%d",&t)!=EOF)    {       while(t--)       {         scanf("%d%d",&n,&m);          for(i=0;i   

#include #include int main() { int n,i,t,j; __int64 a[40][40]; memset(a,0,sizeof(a)); a[1][1]=1; for(i=2;i<=36;i++) for(j=1;j<=i;j++) a[i][j]=a[i-1][j]+a[i][j-1]; t=0; wh

{A} + {B} Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11795    Accepted Submission(s): 4926 Problem Description 给你两个集合，要求{A} + {

#include //#include #include int main() { int n,m,a,i,j,x,y,t; while(scanf("%d%d",&n,&m)!=EOF) { t=0; for(i=1;i<=n;i++) for(j=1;j<=m;j++) { scanf(

#include #include int main() { int n,m,s[60][10],i,j,k,l; double a[60],b[10]; while(scanf("%d%d",&n,&m)!=EOF) { memset(a,0,sizeof(a)); memset(b,0,sizeof(b));

原题链接 #include #include #include bool cmp(int a,int b) { return a>b; } using namespace std; int main() { int t,i; char a[110]; while(scanf("%d",&t)!=EOF) { getchar(); memset(a,0,size