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杭电ACM 1017 A Mathematical Curiosity

neuxxm posted @ 2011年5月20日 00:37 in 杭电ACM 解题报告 with tags 枚举 , 1553 阅读

A Mathematical Curiosity

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10170 Accepted Submission(s): 3078

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

Sample Output

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

#include <stdio.h>
int a2b2[101][101];
int ab[101][101];
int a[101];
int main(){
    int zz, z, n, m, i, j, b, r;
    for (i=1; i<=100; ++i){
        a[i] = i*i;
    }
    for (i=1; i<=100; ++i){
        for (j=1; j<=100; ++j){
            a2b2[i][j] = a[i]+a[j];
            ab[i][j] = i*j;
        }
    }
    b = 0;
    scanf("%d", &zz);
    while (zz-- != 0){
        if (b){
            printf("\n");
        }
        b = 1;
        z = 0;
        for (;;){
            scanf("%d %d", &n, &m);
            if (n==0 && m==0){
                break;
            }
            r = 0;
            for (i=1; i<n; ++i){
                for (j=i+1; j<n; ++j){
                    if ((a2b2[i][j]+m)%ab[i][j] == 0){
                        ++r;
                    }
                }
            }
            printf("Case %d: %d\n", ++z, r);
        }
    }
    return 0;
}

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