A Mathematical Curiosity

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于 2008-05-03 00:14:00 发布

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分类专栏： acm程序设计文章标签： each input output float c

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本文介绍了一种算法，用于解决特定数学问题：给定两个整数n和m，统计满足条件0<a<b<n且(a²+b²+m)/(ab)为整数的整数对(a,b)的数量。输入包含多组n和m的值，输出每组对应的符合条件的整数对数量。

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Given two integers n and m, count the number of pairs of integers (a,b) such that 0 < a < b < n and (a²+b²+m)/(ab) is an integer.

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

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

my c++ code:

#include <iostream>
using namespace std;
int main(){
    int n,m;
    int k = 0;
    cin>>n>>m;
    while(n){
        k++;
        int ans=0;
        float dou,cha;
        int num;
        for (double a = 1; a < n-1; a++){
            for (double b = a+1; b < n; b++){

                dou = (a*a + b*b + m)/(a*b);
                num = (a*a+b*b+m)/(a*b);
                cha = dou - num;
                if (cha==0.0){
                    ans++;
                }
            }
        }
        cout << "Case " << k <<": " << ans << endl;
        cin>>n>>m;
    }

    return 0;
}