2018中国大学生程序设计竞赛 - 网络选拔赛 4 - Find Integer 【费马大定理+构造勾股数】

Find Integer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 6597    Accepted Submission(s): 1852
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 a^n+b^n=c^n.

 

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

题目大意：

已知 n、 a 求解方程 a^n + b^n = c^n;

解题思路：

根据费马大定理 n > 2 时无解；

n == 0 时 无解；

n == 1 时 a， 1， a+1；

n == 2 时，根据已知 a 构造一组勾股数

当 a 为 大于 4 的偶数 2*k 时：  b = k^2 - 1, c = k^2 + 1;

当 a 为 大于 1 的奇数 2*k +1 时： b = 2*k^2 + 2*k, c = 2*n^2 + 2*n + 1;

code：

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <cmath>
 #define mod 1000000007
 #define INF 0x3f3f3f3f
 using namespace std;

 const int MAXN = 1e9;
 int N, a;
 int main()
 {
     int T, b, c, k;
     scanf("%d", &T);
     while(T--)
     {
         scanf("%d%d", &N, &a);
         if(N> || N==) printf("-1 -1\n");
         else if(N==) printf("%d %d\n", , a+);
         if(N==)
         {
             if(a%){
                 k = (a-)/;
                 b = *k*k+*k;
                 c = *k*k+*k+;
                 printf("%d %d\n", b, c);
             }
             else
             {
                 k = a/;
                 b = k*k-;
                 c = k*k+;
                 printf("%d %d\n", b, c);
             }
         }
     }
     return ;
 }