UVA 10692 Huge Mods(指数循环节)

Posted 2020-08-21

指数循环节，由于a ^x = a ^(x % m + phi(m)) (mod m)仅在x >= phi(m)时成立，故应注意要判断

 

//by:Gavin http://www.cnblogs.com/IMGavin/
//指数循环节 递归处理
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 10008, INF = 0x3F3F3F3F;
LL a[N], mod, n;
int phi[N];
void allPhi(int n){
    memset(phi,0,sizeof(phi));
    phi[1]=1;
    for(int i=2;i<=n;i++){
        if(!phi[i]){//则i为素数
            phi[i]=i;
            for(int j=i;j<=n;j+=i){
                    if(!phi[j]){
                        phi[j]=j;
                    }
                phi[j]=phi[j]/i*(i-1);
            }
        }
    }
}
LL PowMod(LL a,LL b,LL MOD){
	LL ret=1;
	while(b){
		if(b&1){
            ret = ret * a % MOD;
		}
		a = a * a % MOD;
		b>>=1;
	}
	return ret;
}

bool check(LL a, LL n, LL m){
    if(a == 1){
        return false;
    }
    LL ans = 1;
    for(int i = 0; i < n; i++){
        ans *= a;
        if(ans >= m){
            return true;
        }
    }
    return false;
}

LL dfs(LL d, LL m, bool &sym){
    if(d == n){
        if(a[d] >= m){
            sym = 1;
        }else{
            sym = 0;
        }
        return a[d] % m;
    }
    bool flag;
    LL p = dfs(d + 1, phi[m], flag);
    if(flag){
        p += phi[m];
    }
    sym = check(a[d], p, m);
    return PowMod(a[d], p, m);
}

int main(){
    allPhi(N - 2);
    int t = 0;
    while(cin >> mod){
        cin >> n;
    	for(int i = 1; i <= n; i++){
    		cin >> a[i];
    	}
    	bool flag;
    	printf("Case #%d: %lld\\n", ++t, dfs(1, mod, flag) % mod);
    }
    return 0;
}