题目链接 Power Tower
题意 给定一个序列,每次给定$l, r$
求$w_{l}^{w_{l+1}^{w_{l+2}^{...^{w_{r}}}}}$ 对m取模的值
根据这个公式
每次递归计算。
因为欧拉函数不断迭代,下降到$1$的级别大概是$log(m)$的,那么对于每一次询问最多需要递归$log(m)$次
注意每次求解欧拉函数的时候要用map存下来,方便以后查询
#include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) #define dec(i, a, b) for (int i(a); i >= (b); --i) typedef long long LL; const int N = 1e5 + 10; int n, q; LL a[N]; LL m; map <LL, LL> f; LL phi(LL n){ if (f.count(n)) return f[n]; LL ans = n, z = n; for (LL i = 2; i * i <= n; ++i){ if (n % i == 0){ ans -= ans / i; while (n % i == 0) n /= i; } } if (n > 1) ans -= ans / n; return f[z] = ans; } LL Pow(LL a, LL b, LL mod){ LL ret = 1; LL fl = a >= mod; for (; b; b >>= 1){ if (b & 1){ ret *= a; if (ret >= mod) fl = 1, ret %= mod; } a *= a; if (a >= mod) a %= mod, fl = 1; } return ret + fl * mod; } LL solve(int l, int r, LL mod){ if (l == r) return a[l]; if (mod == 1) return 1; return Pow(a[l], solve(l + 1, r, phi(mod)), mod); } int main(){ scanf("%d%lld", &n, &m); rep(i, 1, n) scanf("%lld", a + i); scanf("%d", &q); while (q--){ int x, y; scanf("%d%d", &x, &y); printf("%lld\n", solve(x, y, m) % m); } return 0; }
原文地址:https://www.cnblogs.com/cxhscst2/p/8374885.html
时间: 2024-10-07 22:43:37