HDU1017——A Mathematical Curiosity

A Mathematical Curiosity

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37533    Accepted Submission(s): 12003

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

 
解：题意很简单，满足0<a<b<n,并且(a*a+b*b+m)%(a*b)==0就可以了，此时应注意n、m有一个是0就要退出(当时没有注意WA了好几次)。

#include<stdio.h>int main(){int N;scanf("%d",&N);while(N--){int n,m;int count,ca;ca=1;while(scanf("%d%d",&n,&m) && (n ||m)){count=0;for(int a=1;a<n;a++){for(int b=a+1;b<n;b++){if((a*a+b*b+m)%(a*b)==0){count++;}}}printf("Case %d: %d\n",ca++,count);}if(N)printf("\n");}return 0;}