P3934 [Ynoi2016] 炸脖龙 I Solution-优快云博客

本文链接：https://blog.youkuaiyun.com/sblsf/article/details/145310971

Description

给定序列 $a=(a_1,a_2,\cdots,a_n)$ .

有 $m$ 个操作，分以下两种：

$\operatorname{modify}(l,r,k)$ ：对每个 $\in [l,r]$ 执行 $a_i \leftarrow a_i+k$
$\operatorname{query}(l,r,p)$ ：定义 $f(l,l)=a_l$ ， $f(l,r)=a_l^{f(l+1,r)} \;(l < r)$ ，求 $\bmod p$ 。

Limitations

$\le n,m \le 5\times 10^5$
$\le a_i \le 2\times 10^9$
$\le p \le 2\times 10^7$
$\le k \le 2\times 10^9$
$2.5\text{s},512\text{MB}$

Solution

连幂运算不满足结合律，不能线段树维护。
考虑扩展欧拉定理：
$a^b \equiv a^{b \bmod \phi(p)+\phi(p)\times[b \ge \phi(p)]} \pmod p$
我们直接暴力 dfs 遍历每个数求值，由于每个 $p$ 最多递归 $\log p$ 层，因此不会超时。
但是，如何判断 $\ge \phi(p)]$ ？其实不难，跑快速幂时判断一下即可。
然后考虑边界情况:

$p = 1$ 时，答案一定为 $0$
$a_l=1$ 时，答案一定为 $1$

$a$ 用树状数组维护，再筛一下 $\phi(p)$ 即可。

Code

$3.58\text{KB},3.1\text{s},95.71\text{MB}\;\texttt{(in total, C++ 20 with O2) }$

// Problem: P3934 [Ynoi2016] 炸脖龙 I
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3934
// Memory Limit: 512 MB
// Time Limit: 2500 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;

using i64 = long long;
using ui64 = unsigned long long;
using i128 = __int128;
using ui128 = unsigned __int128;
using f4 = float;
using f8 = double;
using f16 = long double;

template<class T>
bool chmax(T &a, const T &b){
	if(a < b){ a = b; return true; }
	return false;
}

template<class T>
bool chmin(T &a, const T &b){
	if(a > b){ a = b; return true; }
	return false;
}

int lowbit(int x){
	return x & -x;
}

template<class T>
struct fenwick{
	int n;
	vector<T> c;
	
	fenwick() {}
	fenwick(int _n): n(_n){
		c.resize(n + 1);
	}
	
	fenwick(const vector<T> &a): n(a.size()){
		c.resize(n + 1);
		for(int i = 1; i <= n; i++){
			c[i] = c[i] + a[i - 1];
			int j = i + lowbit(i);
			if(j <= n) c[j] = c[j] + c[i];
		}
	}
	
	void add(int x, const T& v){
		for(int i = x + 1; i <= n; i += lowbit(i)) c[i] = c[i] + v;
	}
	
	T ask(int x){
		T ans{};
		for(int i = x + 1; i; i -= lowbit(i)) ans = ans + c[i];
		return ans;
	}
	
	T ask(int l, int r){
		return ask(r) - ask(l - 1);
	}
};

using Node = pair<i64, bool>;

Node power(i64 a, i64 b, i64 p) {
    Node res(1, false);
    
    i64 t = a % p;
    if (a >= p) {
        res.second = true;
        res.first %= p;
    }
    
    while (b) {
        if (b & 1) {
            res.first *= t;
        }
        if (res.first >= p) {
            res.second = true;
            res.first %= p;
        }
        
        t *= t;
        if ((b / 2) && t >= p) {
            res.second = true;
            t %= p;
        }
        b >>= 1;
    }
    return res;
}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	int n, m;
	scanf("%d %d", &n, &m);
	
	vector<i64> a(n);
	fenwick<i64> bit(n);
	for (int i = 0; i < n; i++) {
	    scanf("%lld", &a[i]);
	    bit.add(i, a[i] - (i > 0 ? a[i - 1] : 0LL));
	}
	
	vector<bool> np;
	vector<int> pri, phi;
	auto sieve = [&](int n) {
	    np.resize(n + 1, false);
	    phi.resize(n + 1, 0);
	    
	    phi[1] = 1;
	    for (int i = 2; i <= n; i++) {
	        if (!np[i]) {
	            pri.push_back(i);
	            phi[i] = i - 1;
	        }
	        
	        for (int j = 0; j < pri.size() && 1LL * i * pri[j] <= n; j++) {
	            np[i * pri[j]] = true;
	            if (i % pri[j] == 0) {
	                phi[i * pri[j]] = phi[i] * pri[j];
	                break;
	            }
	            phi[i * pri[j]] = phi[i] * (pri[j] - 1);
	        }

	    }
	};
	sieve(2e7);
	
	auto solve = [&](auto &&self, int l, int r, i64 p) -> Node {
        i64 t = bit.ask(l);
        if (p == 1) {
            return Node(0, true);
        }
        if (l == r) {
            return Node(t % p, t >= p);
        }
        if (t == 1) {
            return Node(1, false);
        }
        
        Node mid = self(self, l + 1, r, phi[p]);
        if (mid.second) {
            mid.first += phi[p];
        }
        return power(t, mid.first, p);
    };
    
    for (int i = 0; i < m; i++) {
        int op, l, r;
        i64 p;
        scanf("%d %d %d %lld", &op, &l, &r, &p);
        l--, r--;
        
        if (op == 1) {
            bit.add(l, p);
            bit.add(r + 1, -p);
        }
        else {
            i64 res = solve(solve, l, r, p).first;
            printf("%lld\n", (res % p + p) % p);
        }
    }
	
	return 0;
}