CodeForces 438D :The Child and Sequence 线段树

传送门

题意

给定数列，区间查询和，区间取模，单点修改。

分析

这道题跟花神游历各国类似
首先我们知道，每一个点可以操作的次数是有限的，具体是多少可以自己去证一下，我们只要去维护每一个区间的最大值，如果区间最大值比我取模的数值小，就不需要进行修改了

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
	char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
	while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
struct Node{
	int l,r;
	int mx;
	ll sum;
}tr[N * 4];
int a[N];
int n,m;

void push(int u){
	tr[u].mx = max(tr[u << 1].mx,tr[u << 1 | 1].mx);
	tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}

void build(int u,int l,int r){
	tr[u] = {l,r};
	if(l == r){
		tr[u].mx = a[l];
		tr[u].sum = a[l];
		return;
	}
	int mid = (l + r) >> 1;
	build(u << 1,l,mid),build(u << 1 | 1,mid + 1,r);
	push(u);
}

void modify(int u,int l,int r,int k){
	if(tr[u].mx < k) return;
	if(tr[u].l == tr[u].r){
		tr[u].mx %= k;
		tr[u].sum = tr[u].mx;
		return;
	}
	int mid = (tr[u].l + tr[u].r) >> 1;
	if(l <= mid) modify(u << 1,l,r,k);
	if(r > mid) modify(u << 1 | 1,l,r,k);
	push(u);
}

void modify(int u,int p,int x){
	if(tr[u].l == tr[u].r){
		tr[u].mx = tr[u].sum = x;
		return;
	}
	int mid = (tr[u].l + tr[u].r) >> 1;
	if(p <= mid) modify(u << 1,p,x);
	else modify(u << 1 | 1,p,x);
	push(u);
}

ll query(int u,int l,int r){
	if(tr[u].l >= l && tr[u].r <= r) return tr[u].sum;
	ll res = 0;
	int mid = (tr[u].l + tr[u].r) >> 1;
	if(l <= mid) res = query(u << 1,l,r);
	if(r > mid) res += query(u << 1 | 1,l,r);
	return res;
}

int main() {
	read(n),read(m);
	for(int i = 1;i <= n;i++) read(a[i]);
	build(1,1,n);
	while(m--){
		int op,x,y,z;
		read(op),read(x),read(y);
		if(op ==1){
			dl(query(1,x,y));
		}
		else if(op == 2){
			read(z);
			modify(1,x,y,z);
		}
		else{
			modify(1,x,y);
		}
	}
	return 0;
}