hdu 1017

                     A Mathematical Curiosity

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37621    Accepted Submission(s): 12035

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, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

 

Input

You will be given a number of cases in the input. Each case is specified by a line containing the integers n and m. The end of input is indicated by a case in which n = m = 0. You may assume that 0 < n <= 100.

 

Output

For each case, print the case number as well as the number of pairs (a,b) satisfying the given property. Print the output for each case on one line in the format as shown below.

 

Sample Input

110 120 330 40 0

 

Sample Output

Case 1: 2Case 2: 4Case 3: 5

题意很简单：给你数n,从里面找出两个数a,b，0 < a < b < n and (a^2+b^2 +m)%(ab)==0

这本来是一道很简单的题目，但当时有很多同学是输出格式错误，算的上水坑题。

哈哈，因为他们可能忽略了这句话“The output format consists of N output blocks. There is a blank line between output blocks.”

大意就是在两者之间需要输出一个空行，其余的应该就没问题了

代码：

#include <iostream>#include<cstdio>using namespace std;int main(){    int N;    scanf("%d",&N);    while(N--)    {        int cas=1;        int x,y;        while(~scanf("%d%d",&x,&y)&&(x+y))        {            int cnt=0;            for(int i=1;i<x;i++){               for(int j=i+1;j<x;j++) {                if((i*i+j*j+y)%(i*j)==0)                    cnt++;               }            }            printf("Case %d: %d\n",cas++,cnt);        }        if(N)            printf("\n");    }    return 0;}