【bzoj3884】上帝与集合的正确用法

【题目大意】

求2^(2^(2^(2^(2^...)))) mod p的值

【题解】

这是道神题。

来自PoPoQQQ大爷的题解

 1 /*************
 2   bzoj 3884
 3   by chty
 4   2016.11.4
 5 *************/
 6 #include<iostream>
 7 #include<cstdio>
 8 #include<cstring>
 9 #include<cstdlib>
10 #include<ctime>
11 #include<cmath>
12 #include<algorithm>
13 using namespace std;
14 typedef long long ll;
15 inline ll read()
16 {
17     ll x=0,f=1;  char ch=getchar();
18     while(!isdigit(ch))  {if(ch=='-')  f=-1;  ch=getchar();}
19     while(isdigit(ch))  {x=x*10+ch-'0';  ch=getchar();}
20     return x*f;
21 }
22 ll fast(ll a,ll b,ll mod){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b/=2;a=a*a%mod;}return ans;}
23 ll get(ll x)
24 {
25     ll sum=x;
26     for(ll i=2;i*i<=x;i++)
27         if(x%i==0)
28         {
29             sum=sum/i*(i-1);
30             while(x%i==0)  x/=i;
31         }
32     if(x>1)  sum=sum/x*(x-1);
33     return sum;
34 }
35 ll solve(ll p)
36 {
37     if(p==1)  return 0;
38     ll temp=0;
39     while(~p&1)  p/=2,temp++;
40     ll phi=get(p);
41     ll re=solve(phi);
42     re=(re+phi-temp%phi)%phi;
43     re=(fast(2,re,p))%p;
44     return re<<temp;
45 }
46 int main()
47 {
48     freopen("cin.in","r",stdin);
49     freopen("cout.out","w",stdout);
50     ll T=read();
51     while(T--)
52     {
53         ll p=read(); 
54         printf("%lld\n",solve(p));
55     }
56     return 0;
57 }

 

posted @ 2016-11-04 16:22  chty  阅读(205)  评论(0)  编辑  收藏  举报