题解 - [CodeForces 906 D] Power Tower

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题目链接

原始题面

time limit per test: 4.5 seconds
memory limit per test: 256 megabytes
input: standard input
output: standard output

Description

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 \(w_1, ..., w_m\) value of power will be

\[ w_1^{w_2^{\cdot^{\cdot^{\cdot^{w_m}}}}} \]

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 \(n\) (\(1 ≤ n ≤ 10^5\)) and \(m\) (\(1 ≤ m ≤ 10^9\))

Second line of input contains \(n\) integers \(w_k\) (\(1 ≤ w_k ≤ 10^9\)) which is the power of rocks that priests have

Third line of input contains single integer \(q\) (\(1 ≤ q ≤ 10^5\)) which is amount of queries from priests to you

kth 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

1
2
3
4
5
6
7
8
9
10
11
6 1000000000
1 2 2 3 3 3
8
1 1
1 6
2 2
2 3
2 4
4 4
4 5
4 6

Output

1
2
3
4
5
6
7
8
1
1
2
4
256
3
27
597484987

Note

\(3^{27} = 7625597484987\)

题意简述

给出 \(n,m,w_1,w_2,\dots,w_n\), 有 \(q\) 组询问，每次给出 \(l,r\), 问

\[ w_l^{w_{l+1}^{\cdot^{\cdot^{\cdot^{w_r}}}}}\bmod m \]

解题思路

由扩展 Euler 定理

\[ a^b\equiv\begin{cases} a^{b\bmod\varphi(m)+\varphi(m)},&(a,m)>1,b>\varphi(m)\\ a^{b\bmod\varphi(m)},&\texttt{otherwise}\\ \end{cases}\pmod m \]

递归求解即可

复杂度

\[ \Theta\left(\log m+\sum_{i=1}^q\min\{r_i-l_i+1,\log m\}\right)\implies O(q\log m) \]

代码参考

Show code

CodeForces_906Dview raw
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/*
 * @Author: Tifa
 * @Description: From <https://github.com/Tiphereth-A/CP-archives>
 * !!! ATTENEION: All the context below is licensed under a 
 *  GNU Affero General Public License, Version 3.
 *  See <https://www.gnu.org/licenses/agpl-3.0.txt>.
 */
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
unordered_map<int, int> phi;
int calc_phi(int p) {
  int ans = p;
  for (int i = 2; i <= sqrt(p); ++i)
    if (p % i == 0) {
      ans = ans / i * (i - 1);
      while (p % i == 0) p /= i;
    }
  if (p > 1) ans = ans / p * (p - 1);
  return ans;
}
i64 qpow(i64 a, i64 b, i64 mod) {
  i64 res = 1;
  a > mod ? (a %= mod) += mod : 0;
  for (; b; b >>= 1, (a *= a) > mod ? (a %= mod) += mod : 0)
    if (b & 1) (res *= a) > mod ? (res %= mod) += mod : 0;
  return res;
}
int w[N];
i64 dfs(int l, int r, int x, i64 p) {
  if (x > r || p == 1) return 1;
  return qpow(w[x], dfs(l, r, x + 1, phi[p]), p);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", w + i);
  int _m = m;
  phi[1] = 1;
  while (_m > 1) {
    phi[_m] = calc_phi(_m);
    _m = phi[_m];
  }
  int kase;
  scanf("%d", &kase);
  while (kase--) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%lld\n", dfs(l, r, l, m) % m);
  __end_kase:;
  }
  return 0;
}