HDOJ 1017 A Mathematical Curiosity
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A Mathematical Curiosity
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21549 Accepted Submission(s): 6736
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
110 120 330 40 0
Sample Output
Case 1: 2Case 2: 4Case 3: 5我反正没理解这个测试数据的1是什么意思 还有题目不是写的m n 同时为0才结束么
#include <stdio.h>
#include <iostream>
using namespace std;
int kaka(int a,int b)
{int cnt=0,i,j;
for (i=1;i<=a-2;i++)
for (j=i+1;j<=a-1;j++)
if ((i*i+j*j+b)%(i*j)==0)
cnt++;
return cnt;
}
int main()
{
int t,n,m,cmp1,i;
cin>>t;
for (i=0;i<t;i++)
{ cmp1=1;
while (cin>>n>>m&&(n!=0||m!=0))
{
cout<<"Case "<<cmp1++<<": "<<kaka(n,m)<<endl;
}
if (i!=t-1)
cout<<endl;
}
return 0;
}
#include <iostream>
using namespace std;
int kaka(int a,int b)
{int cnt=0,i,j;
for (i=1;i<=a-2;i++)
for (j=i+1;j<=a-1;j++)
if ((i*i+j*j+b)%(i*j)==0)
cnt++;
return cnt;
}
int main()
{
int t,n,m,cmp1,i;
cin>>t;
for (i=0;i<t;i++)
{ cmp1=1;
while (cin>>n>>m&&(n!=0||m!=0))
{
cout<<"Case "<<cmp1++<<": "<<kaka(n,m)<<endl;
}
if (i!=t-1)
cout<<endl;
}
return 0;
}
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