Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 6597 Accepted Submission(s): 1852
Special Judge
定理：
费马大定理
a^n+b^n = c^n 若n大于2则无正整数解
只需要特判下0，1 然后找下2的解就可以
#include<bits/stdc++.h> using namespace std; typedef long long ll; int main() { int t,n,a; scanf("%d",&t); while(t--) { scanf("%d %d",&n,&a); if(n==0) { printf("-1 -1\n"); } if(n==1) { printf("1 %d\n",a+1); } if(n==2) { if(a%2==0) { printf("%d %d\n",(a*a/2-2)/2,(a*a/2+2)/2); } else { printf("%d %d\n",(a*a-1)/2,(a*a+1)/2); } } if(n>=3) { printf("-1 -1\n"); } } return 0; }
原文：https://www.cnblogs.com/hao-tian/p/9536405.html