HDU 6441(費馬大定理+奇偶數列法)
阿新 • • 發佈:2019-01-02
思路:由費馬大定理知a^n+b^n=c^n當n>2時無整數解,所以n==0和n>2時輸出-1 -1,n==1時輸出1,a+1,n==2時,由奇偶數列法
#include<bits/stdc++.h> using namespace std; #define inf 0x3f3f3f3f #define ll long long const int maxn=200005; const double eps=1e-8; const double PI = acos(-1.0); #define lowbit(x) (x&(-x)) int main() { int t; cin>>t; while(t--) { int n,a; scanf("%d %d",&n,&a); if(n>2||n==0) { printf("-1 -1\n"); } else if(n==1) { printf("1 %d\n",a+1); } else if(n==2) { if(a%2) { n=(a-1)/2; printf("%d %d\n",n*n+(n+1)*(n+1)-1,n*n+(n+1)*(n+1)); } else { n=a/2; printf("%d %d\n",n*n-1,n*n+1); } } } return 0; }