UVALive

由 费马小定理可知：

a ^ b % m  = a ^ ( b % (m-1) ) % m ;

（费马小定理：m是质数时  a ^ (m-1) % m = 1）

所以对于本题 枚举 a 的 1 - p  次幂对 p 取模，

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;typedef long long ll;const int maxn = 10 + 7, INF = 0x3f3f3f3f, mod = 1e9 + 7;ll p, a, b, x;bool solve(ll n) {    //cout << n << "  ";    ll ans = 1;    ll t = a;    while(n) {        if(n & 1) ans *= t, ans %= p;        t = t * t; t %= p;        n /= 2;    }    //cout << ans << " == " << endl;    if(ans == b) return true;    else return false;}int main() {    ios::sync_with_stdio(0);    cin >> p;    while(cin >> a && a) {        cin >> b;        int f = 0;        for(ll i = 1; i < p; ++i) {            if(solve(i)) { cout << i << endl; f = 1; break;}        }        if(!f) cout << 0 << endl;    }    return 0;}