Find Integer(费马大定理+勾股数)_算法 find integer-优快云博客

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博客围绕整数查找问题展开，给定整数n和a，需找出满足an+bn=cn的b和c。介绍了该问题的输入输出格式及样例，解题思路运用费马大定理和勾股数知识，当n>2时无正整数解，还给出了奇数和偶数情况下的勾股数公式。

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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

2018中国大学生程序设计竞赛 - 网络选拔赛

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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;
}