ACM HDU 1017 A Mathematical Curiosity
A Mathematical Curiosity
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17031 Accepted Submission(s): 5306
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.
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
Source
Recommend
JGShining
#include<stdio.h> int main() { int n, m, times, count = 0, i, j, flag = 1; scanf("%d", ×); while(times--) { while(1) { scanf("%d%d", &n, &m); if(!n && !m) break; for(i=1; i<n; ++i) for(j=1; j<i; ++j) if((i*i+j*j+m)%(i*j) == 0) count++; printf("Case %d: %d\n", flag++, count); count = 0; } flag = 1; if(times != 0) printf("\n"); } return 0; }
解题报告:
这题明显的水题,之所以会PE,我不会承认是我的错,只是把N跟第一组 input 之间的 Blank 忽略掉就没问题了。
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作者:雪影蓝枫
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