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
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
暴力枚举
看懂题意暴力枚举就好~~
单纯的枚举题,注意每两个输入之间需要输出一次\n#include<iostream>
using namespace std;
int main()
{
int t, n, m, i, j;
cin>>t;
while(t--)
{
int num = 1;
while(cin>>n>>m && n||m)
{
int cnt = 0;
for(i = 1;i < n;i++)
{
for(j = i+1;j < n;j++)
{
if((i * i + j * j + m)%(i * j) == 0)
cnt++;
}
}
printf("Case %d: %d\n",num,cnt);
num++;
}
if(t)printf("\n");
}
return 0;
}
本文探讨了一道涉及数学和编程的问题:对于给定的整数n和m,寻找所有符合条件0<a<b<n且(a²+b²+m)/(ab)为整数的整数对(a,b)的数量。通过枚举的方法解决此问题,并给出了具体的实现代码。
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