首页 > 代码库 > HDU1017
HDU1017
A Mathematical Curiosity Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 24900 Accepted Submission(s): 7846 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 1 10 1 20 3 30 4 0 0 Sample Output Case 1: 2 Case 2: 4 Case 3: 5
//Problem:hdu1017 //Data:2011/10/30 #include <iostream> using namespace std; int main() { //freopen("E:\\in.txt","r",stdin); int N; cin >> N; while(N--) { int cas = 1; int n,m; int a,b; while(cin>>n>>m, n||m) { int count = 0; for(a=1; a<n; a++) { for(b=a+1; b<n; b++) { if((a*a + b*b + m) % (a*b) == 0) count++; } } cout <<"Case "<<cas++<<": "<<count<<endl; } if(N) cout << endl; } return 0; }
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。