BZOJ 3884 (欧拉函数)

题目链接：点击这里

题意：求2222....%p.

设p=2n+q,那么2222....%p=2n(2(2222....−n)%q)=2n(2(2222....−n)%ϕ(q)%q)

然后一直递归下去直到ϕ(q)=1就可以回溯了。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <map>
using namespace std;
#define maxn 10050000
#define mod 1000000007

bool vis[maxn];
int phi[maxn], prime[maxn], cnt;

void init () {
cnt = 0;
memset (vis, 0, sizeof vis);
phi[1] = 1;
for (int i = 2; i < maxn; i++) {
if (!vis[i]) {
prime[cnt++] = i;
phi[i] = i-1;
}
for (int j = 0; j < cnt; j++) {
if (1LL*i*prime[j] >= maxn) break;
vis[i*prime[j]] = 1;
if (i%prime[j] == 0) {
phi[i*prime[j]] = phi[i] * prime[j];
break;
}
else {
phi[i*prime[j]] = phi[i] * (prime[j]-1);
}
}
}
}

int p;

long long qpow (long long b, int p) {
if (b == 0)
return 1%p;
long long ans = qpow (b>>1, p);
ans = ans*ans%p;
if (b&1) ans = ans*2%p;
return ans;
}

long long dfs (int cur) {
if (cur == 1)
return 0;
int n = 0;
while (!(cur&1)) n++, cur >>= 1;
long long ans = dfs (phi[cur]);
ans = (ans-n%phi[cur]+phi[cur])%phi[cur];
ans = qpow (ans, cur);
return ans*(1<<n);
}

int main () {
init ();
int t;
scanf ("%d", &t);
while (t--) {
scanf ("%d", &p);
printf ("%lld\n", dfs (p));
}
return 0;
}