杭电OJ_DIY_YTW2_1001 A Mathematical Curiosity
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.
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
Source
East Central North America 1999, Practice
题目分析：题意是说，第一行输入一个数N，分N模块进行输入输出，在模块中输入一组（n，m），使它们同时满足0 < a < b < n和 (a^2+b^2 +m)/(ab)是整数的解的个数。
注意：模块之间的输入输出要有一个空行，还有输入时逻辑运算的应用。
#include<iostream>
using namespace std;
int main()
{
int a,b,m,n,num,i,s,N;
cin>>N;
for(i=0;i<N;i++)
{
s=1;
while(cin>>n>>m,n||m)
{
num=0;
for(a=1;a<n;a++)
{
for(b=a+1;b<n;b++)
{
if((a*a+b*b+m)%(a*b)==0)
num++;
}
}
cout<<"Case "<<s<<": "<<num<<endl;
s++;
}
if(i!=N-1)
cout<<endl; //至关重要，各模块间有一行是空行。
}
return 0;
}
