【bzoj3884】 上帝与集合的正确用法

http://www.lydsy.com/JudgeOnline/problem.php?id=3884 (题目链接)

题意

  求这里写图片描述

Solution

  解决的关键:

  当${n>φ(p)}$,有$${a^n≡a^{n\%φ(p)+φ(p)}~(mod~p)}$$

  然后递归log(p)次就会出解:http://blog.csdn.net/skywalkert/article/details/43955611

细节

代码

// bzoj3884
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#define LL long long
#define inf 2147483640
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
 
const int maxn=10000010;
int phi[maxn],vis[maxn],p[maxn];

void calphi() {
	phi[1]=1;
	for (int i=2;i<maxn;i++) {
		if (!vis[i]) {p[++p[0]]=i;phi[i]=i-1;}
		for (int j=1;j<=p[0];j++) {
			if (p[j]*i>maxn) break;
			vis[p[j]*i]=1;
			if (i%p[j]==0) {phi[p[j]*i]=phi[i]*p[j];break;}
			else phi[p[j]*i]=phi[p[j]]*phi[i];
		}
	}
}
int power(int a,int b,int c) {
	int res=1;
	while (b) {
		if (b&1) res=(LL)res*a%c;
		b>>=1;a=(LL)a*a%c;
	}
	return res;
}
int solve(int p) {
	if (p==1) return 0;
	int res=solve(phi[p])+phi[p];
	return power(2,res,p);
}
int main() {
	calphi();
	int T,P;scanf("%d",&T);
	while (T--) {
		scanf("%d",&P);
		printf("%d\n",solve(P));
	}
    return 0;
}

  

posted @ 2016-12-13 09:12  MashiroSky  阅读(...)  评论(...编辑  收藏