51nod 1135 原根（求素数最小原根）

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题目链接：
51nod 1135 原根
题意：
求素数p的最小原根。

#include <stdio.h>#include <algorithm>#include <cmath>#include <iostream>#include <string.h>#include <cstring>using namespace std;const int MAX_N = 1000010;typedef long long ll;int prime_cnt, factor_cnt, p;int vis[MAX_N], prime[MAX_N], factor[MAX_N];void GetPrime(){    prime_cnt = 0;    memset(vis, 0, sizeof(vis));    for(int i = 2; i < MAX_N; i++){        if(!vis[i]){            prime[prime_cnt++] = i;            for(int j = 0; j < prime_cnt && prime[j] * i < MAX_N; j++){                vis[i * prime[j]] = 1;                if(i % prime[j] == 0) break;            }        }    }}void Factor(int x){    factor_cnt = 0;    int t = (int) sqrt(x + 0.5);    for(int i = 0; prime[i] <= t; i++){        if(x % prime[i] == 0){            factor[factor_cnt++] = prime[i];            while(x % prime[i] == 0) x /= prime[i];        }    }    if(x > 1) factor[factor_cnt++] = x;}ll quick_pow(ll n, ll m, ll mod){    ll res = 1, tmp = n % mod;    while(m){        if(m & 1) res = res * tmp % mod;        tmp = tmp * tmp % mod;        m >>= 1;    }    return res;}void solve(){    Factor(p - 1);    for(int g = 2; g < p; g++){        bool flag = true;        for(int i = 0; i < factor_cnt; i++){            int t = (p - 1) / factor[i];            if(quick_pow((ll)g, (ll)t, (ll)p) == 1){                flag = false;                break;            }        }        if(flag){            cout << g << endl;            break;        }    }}int main(){    GetPrime();    while(cin >> p){        solve();    }    return 0;}

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