方法:枚举 做这道题,在第十行的位置WA了很多次,(n || m)不能写成(n + m),m可能是负数,一直没注意到,导致不停的WA....... #include <stdio.h> int main() { int T,n,m,a,b,ans,p = 0; scanf("%d",&T); while(T--) { if(p ++) printf("\n"); int num = 1; while(scanf("%d%d",…

A Mathematical Curiosity Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 31177 Accepted Submission(s): 9988 Problem Description Given two integers n and m, count the number of pairs of integers (a…

A Mathematical Curiosity Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 87   Accepted Submission(s) : 23 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Given two in…

A Mathematical Curiosity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41995    Accepted Submission(s): 13502 Problem Description Given two integers n and m, count the number of pairs of integ…

A Mathematical Curiosity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25871    Accepted Submission(s): 8174 Problem Description Given two integers n and m, count the number of pairs of integ…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1017 A Mathematical Curiosity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47658    Accepted Submission(s): 15285 Problem Description Given two…

//2014.10.17    01:19 //题意: //先输入一个数N,然后分块输入,每块输入每次2个数,n,m,直到n,m同一时候为零时  //结束,当a和b满足题目要求时那么这对a和b就是一组 //注意: //每一块的输出中间有一个回车 A Mathematical Curiosity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s…

Problem Description Given two integers n and m, count the number of pairs of integers (a,b) such that 0 < a < b < n and (a^2+b^2 +m)/(ab) is an integer. This problem contains multiple test cases! The first line of a multiple input is an integer N…

http://acm.hdu.edu.cn/showproblem.php?pid=1017 # include <stdio.h> int find(int n, int m) { int count = 0; for(int i = 1; i < n; i++) { for(int j = i + 1; j < n; j++) { if(((i * i) + (j * j) + m) % (i * j) == 0) count++; } } return count; } in…

题目简单,格式酸爽.. #include <iostream> using namespace std; int ans,n,m,p; int main() { cin>>p; while(p--) { ; while(~scanf("%d%d",&n,&m)&&(n+m)) { ans=; ;i<n;i++) { ;j<n;j++) { ) ans++; } } printf("Case %d: %d\n&…