HDU 1017 A Mathematical Curiosity 水题 格式控制

A Mathematical Curiosity

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38898    Accepted Submission(s): 12464

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

 

Source

East Central North America 1999, Practice

 

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 题意很明确，就是求出来 0 < a < b < n and (a^2+b^2 +m)/(ab) is an intege

求出来a，b的个数。

题目很水，就是输出格式得控制一下，每两个n之间价格空格！！！

ac代码：

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