HDOJ 1017 HDU 1017 A Mathematical Curiosity ACM 1017 IN HDU

MiYuオリジナル、転帖は明記してください:転載は______________白い家 から
 
タイトルアドレス:http://acm.hdu.edu.cn/showproblem.php?pid=1017
タイトルの説明:
A Mathematical CuriosityTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8614    Accepted Submission(s): 2592
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
      
        

 
标题分析:无事水一水,へへへ....初心者は出力に注意すればいいだけです.暴力を振るったことがあって、技巧のコードは以下の通りです:*MiYuオリジナル、転帖は明記してください:転載して_____白い家http://www.cnblog.com/MiYuAuthor By : MiYuTest      : 1Program   : 1017*/#include #include using namespace std;int main (){    int T;    cin >> T;    for ( int k = 1; k <= T;++ k ){           int N, M, ca = 1, cnt = 0;           if ( k != 1 ) putchar ( 10 );           while ( cin >> N >> M, N || M ){     cnt = 0;                  for ( int i = 1; i < N;++ i ){                       for ( int j = i + 1; j < N;++ j ){                            if ( ( i * i + j * j + M ) % ( i * j ) == 0 )                                     cnt++;                       }                      }                     printf ( "Case %d: %d",ca++, cnt );              }          }     return 0;}