hdu 1017

A Mathematical Curiosity

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38772 Accepted Submission(s): 12422

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
1

10 1
20 3
30 4
0 0

Sample Output
Case 1: 2
Case 2: 4
Case 3: 5

水题

#include <stdio.h>int name(int n,int m){    int e=0;    int a=1,b;    for(;a<n;a++)    {        for(b=a+1;b<n;b++)        {            if((a*a+b*b+m) %(a*b)==0)            {                e++;            }        }    }    return e;}int main(){    int num;    int n,m;    int c=0;    scanf("%d",&num);    while(num--)    {        int sort=1;        while(scanf("%d%d",&n,&m)!= EOF){            if(n==0&&m==0) break;            c=name(n,m);            printf("Case %d: %d\n",sort,c);            sort++;        }        if(num != 0)            printf("\n");    }    return 0;}

当我写到`while(scanf(“%d%d”,&n,&m)!= EOF){ 的时候我写成了while(scanf(“%d%d”,&n,&m)!= EOF,n&&m) 了，然后就一直WA，才发现不能这么写，好吧，然后我就写了if语句，哎呀，好吧就这样。