AcWing 243 一个简单的整数问题2 (分块做法)

https://www.acwing.com/problem/content/244/

分块入门练习题
大块维护，小块朴素

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;

const int maxn = 100010;

int n, m, t;
int L[maxn], R[maxn], pos[maxn];
ll a[maxn], add[maxn], sum[maxn];

char s[10];

void modify(int l, int r, ll d){
	int p = pos[l], q = pos[r];
	if(p == q){
		for(int i = l; i <= r; ++i) a[i] += d;
		sum[p] += 1ll * (r - l + 1) * d;
	} else{
		for(int i = p + 1; i <= q - 1; ++i) add[i] += d;
		for(int i = l; i <= R[p]; ++i) a[i] += d;
		sum[p] += 1ll * d * (R[p] - l + 1);
		for(int i = L[q]; i <= r; ++i) a[i] += d;
		sum[q] += 1ll * d * (r - L[q] + 1);
	}
}

ll query(int l, int r){
	ll res = 0;
	int p = pos[l], q = pos[r];
	if(p == q){
		for(int i = l; i <= r; ++i) res += a[i];
		res += 1ll * (r - l + 1) * add[p];
	} else{
		for(int i = p + 1; i <= q - 1; ++i) res += sum[i] + 1ll * add[i] * (R[i] - L[i] + 1);
		for(int i = l ; i <= R[p] ; ++i) res += a[i];
		res += 1ll * add[p] * (R[p] - l + 1);
		for(int i = L[q] ; i <= r ; ++i) res += a[i];
		res += 1ll * add[q] * (r - L[q] + 1);
	}
	return res;
}

ll read(){ ll s=0,f=1; char ch=getchar(); while(ch<'0' || ch>'9'){ if(ch=='-') f=-1; ch=getchar(); } while(ch>='0' && ch<='9'){ s=s*10+ch-'0'; ch=getchar(); } return s*f; }

int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; ++i) scanf("%lld",&a[i]);
	// 分块 
	t = sqrt(n);
	for(int i = 1; i <= t; ++i){
		L[i] = (i - 1) * t + 1;
		R[i] = i * t;
	} 
	if(R[t] < n){
		++t; L[t] = R[t - 1] + 1; R[t] = n;
	}
	// 预处理
	for(int i = 1; i <= t; ++i){
		for(int j = L[i]; j <= R[i]; ++j){
			pos[j] = i;
			sum[i] += a[j];
		}
	} 
	
	int l, r, d;
	for(int i = 1; i <= m; ++i){
		scanf("%s%d%d", s, &l, &r);
		if(s[0] == 'C'){
			scanf("%lld", &d);
			modify(l, r, d);
		} else{
			printf("%lld\n", query(l, r));
		}
	}
	
	return 0;
}

标签：ch,分块,int,ll,pos,maxn,243,include,AcWing
来源： https://www.cnblogs.com/tuchen/p/13961079.html