Find Integer
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1029 Accepted Submission(s): 251
Special Judge
Problem Description
people in USSS love math very much, and there is a famous math problem .
give you two integers n,a,you are required to find 2 integers b,c such that an+bn=cn.
Input
one line contains one integer T;(1≤T≤1000000)
next T lines contains two integers n,a;(0≤n≤1000,000,000,3≤a≤40000)
Output
print two integers b,c if b,c exits;(1≤b,c≤1000,000,000);
else print two integers -1 -1 instead.
Sample Input
1 2 3
Sample Output
4 5
Source
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思路:
费马大定理+勾股数
费马大定理:
当整数n>2时,关于x,y,z的方程
没有正整数解。
勾股数:
///n为奇数有勾股数(2*n+1,2*n*n+2*n,2*n*n+2*n+1);n为偶数有勾股数(2*n,n*n-1,n*n+1)。
代码:
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t,n,a,b,c;
scanf("%d",&t);
while(t--){
int flag=0;
scanf("%d%d",&n,&a);
if(n>2 || n==0)
{
printf("-1 -1\n");
continue;
}
if(n==1)
{
b=1;c=a+b;
}
else{
///n为奇数有勾股数(2*n+1,2*n*n+2*n,2*n*n+2*n+1);n为偶数有勾股数(2*n,n*n-1,n*n+1)。
if(a%2==1){
int lala=(a-1)/2;
b=2*lala*lala+2*lala;
c=2*lala*lala+2*lala+1;
}
else{
int lala=a/2;
b=lala*lala-1;
c=lala*lala+1;
}
}
printf("%d %d\n",b,c);
}
return 0;
}