Codeforces 906 D Power Tower

Discription

Priests of the Quetzalcoatl cult want to build a tower to represent a power of their god. Tower is usually made of power-charged rocks. It is built with the help of rare magic by levitating the current top of tower and adding rocks at its bottom. If top, which is built from k?-?1 rocks, possesses power p and we want to add the rock charged with power w_k then value of power of a new tower will be {w_k}^p.

Rocks are added from the last to the first. That is for sequence w₁,?...,?w_m value of power will be

After tower is built, its power may be extremely large. But still priests want to get some information about it, namely they want to know a number called cumulative power which is the true value of power taken modulo m. Priests have n rocks numbered from 1 to n. They ask you to calculate which value of cumulative power will the tower possess if they will build it from rocks numbered l,?l?+?1,?...,?r.

Input

First line of input contains two integers n (1?≤?n?≤?10⁵) and m (1?≤?m?≤?10⁹).

Second line of input contains n integers w_k (1?≤?w_k?≤?10⁹) which is the power of rocks that priests have.

Third line of input contains single integer q (1?≤?q?≤?10⁵) which is amount of queries from priests to you.

k^th of next q lines contains two integers l_k and r_k (1?≤?l_k?≤?r_k?≤?n).

Output

Output q integers. k-th of them must be the amount of cumulative power the tower will have if is built from rocks l_k,?l_k?+?1,?...,?r_k.

Example

Input

6 10000000001 2 2 3 3 381 11 62 22 32 44 44 54 6

Output

1124256327597484987

Note

3²⁷?=?7625597484987

首先你得需要知道一个定理：

当x>φ(p)时，a^x  mod p=a^(x mod φ(p)  +φ(p))  mod p。

因为φ(x)迭代不超过log(x)次就成1了(考虑2这个质因子)，所以我们直接暴力迭代就行了。

由于我们并不知道下一层的值(也就是这一层的次数)是否大于φ(p)，所以%p改成%2p就行了。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<map>
#define ll long long
#define maxn 100005
using namespace std;
ll a[maxn],n,mod[105],l,r,q,tot=0;

inline ll MO(ll x,ll y){
    return x>y?x%y+y:x;
}

inline ll ksm(ll x,ll y,const ll ha){
    ll an=1;
    for(;y;y>>=1,x=MO(x*x,ha)) if(y&1) an=MO(an*x,ha);
    return an;
}

inline ll phi(ll x){
    int tp=sqrt(x+0.5),y=1;
    for(int i=2;i<=tp;i++) if(!(x%i)){
        x/=i,y*=i-1;
        while(!(x%i)) x/=i,y*=i;
        if(x==1) break;
    }
    if(x!=1) y*=x-1;
    return y;
}

ll solve(int now,ll mo){
    if(now==r||mo==1) return MO(a[now],mo);
    else return ksm(a[now],solve(now+1,mod[now-l+1]),mo);
}

int main(){
    scanf("%lld%lld",&n,mod);
    while(mod[tot]!=1) mod[tot+1]=phi(mod[tot]),tot++;

    for(int i=1;i<=n;i++) scanf("%lld",a+i);
    scanf("%lld",&q);
    while(q--){
        scanf("%lld%lld",&l,&r);
        ll ans=solve(l,mod[0]);
        if(ans>=mod[0]) ans-=mod[0];
        printf("%lld\n",ans);
    }

    return 0;
}

原文地址：https://www.cnblogs.com/JYYHH/p/8384375.html