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英文不好,一开始没读懂题,后来发现就是a的p次方模p为a&&p为非素数为yes,否则为no#include <iostream> /*快速幂取模*/ using namespace std; typedef long long ll; ll f(ll x,ll y) { ll base=x,r=1,mo=y; while(y) { if(y&1) r=(r*base)%mo; base=(base*base)%mo; y>>=1; } return r; } bool is_pri(ll x) { for(int i=2;i*i<=x;i++) { if(x%i==0) return false; } return true; } int main() { ll p,a; while(cin >> p >> a,p||a) { if(f(a,p)==a&&!is_pri(p)) cout << "yes" << endl; else cout << "no" << endl; } return 0; } Followed by: Post your reply here: |
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