A Simple Problem with Integers

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博客围绕N个整数展开，需处理两种操作：一是给给定区间内每个数加上指定值，二是查询给定区间内数字之和。输入包含N、Q及初始值，操作指令有两种格式，输出需按顺序给出查询结果，同时提醒注意int越界要用long long类型。

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You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

Hint

The sums may exceed the range of 32-bit integers.

注意：int会越界，需要用long long类型

#include<iostream>
#include<string>
#include<cstring>
#include<string.h>
#include<cstdio>
using namespace std;
typedef long long ll;
const ll maxn=100000;
ll s[4*maxn+100],col[4*maxn+100];//懒惰标记 
void up(int p){
	s[p]=s[p*2]+s[(p*2)+1];
}
void down(int p,int l,int r){//延迟标记 
	if(col[p]){
		int mid=(l+r)/2;
		s[p*2]+=col[p]*(mid-l+1);
		s[(p*2)+1]+=col[p]*(r-mid);
		col[p*2]+=col[p];
		col[(p*2)+1]+=col[p];
		col[p]=0;
	}
}
void modify(int p,int l,int r,int x,int y,int c){
	if(x<=l&&y>=r){
		s[p]+=(r-l+1)*c;
		col[p]+=c;
		return;
	}
	down(p,l,r);
	int mid=(l+r)/2;
	if(x<=mid){
		modify(p*2,l,mid,x,y,c);
	}
	if(y>mid){
		modify(p*2+1,mid+1,r,x,y,c);
	}
	up(p);
}
ll query(int p,int l,int r,int x,int y){
	if(x<=l&&y>=r){
		return s[p];
	}
	down(p,l,r);
	int mid=(l+r)/2;
	ll res=0;
	if(x<=mid){
		res+=query(p*2,l,mid,x,y);
	}
	if(y>mid){
		res+=query(p*2+1,mid+1,r,x,y);
	}
	return res;
}
int main(){
	int n,q,x;
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;i++){
		scanf("%d",&x);
		modify(1,1,n,i,i,x);
	}
	int a,b,c;
	char ch;
	for(int i=1;i<=q;i++){
		cin>>ch;
		if(ch=='Q'){
			scanf("%d%d",&a,&b);
				printf("%lld\n",query(1,1,n,a,b));
			}else{
			scanf("%d%d%d",&a,&b,&c);
				modify(1,1,n,a,b,c);
			}
	}
	return 0;
}