poj 3468

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 90736		Accepted: 28268
Case Time Limit: 2000MS

Description

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

Input

The first linecontains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means queryingthe sum of A_a, A_a+1, ... , A_b.

Output

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

SampleInput

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

SampleOutput

4

55

9

15

Hint

The sumsmay exceed(超过) the range of 32-bit integers(整数).

Source

POJMonthly--2007.11.25, Yang Yi

 注意：数组要用long long

 线段树解法

Problem: 3468

User: ksq2013

Memory: 4776K

Time: 2532MS

Language: G++

Result: Accepted

#include<cstdio>
#include<cstdlib>
#include<iostream>
using namespace std;
int n,q;
long long col[400001],sum[400001];
void pushup(int k)
{sum[k]=sum[k<<1]+sum[k<<1|1];}
void pushdown(int k,int m)
{
	if(col[k]){
		col[k<<1]+=col[k];
		col[k<<1|1]+=col[k];
		sum[k<<1]+=col[k]*(m-(m>>1));
		sum[k<<1|1]+=col[k]*(m>>1);
		col[k]=0;
	}
}
void build(int s,int t,int k)
{
	if(!(s^t)){
		scanf("%lld",&sum[k]);
		return;
	}int m=(s+t)>>1;
	build(s,m,k<<1);
	build(m+1,t,k<<1|1);
	pushup(k);
}
void update(int s,int t,int k,int l,int r,int c)
{
	if(l<=s&&t<=r){
		col[k]+=c;
		sum[k]+=c*(t-s+1);
		return;
	}pushdown(k,t-s+1);
	int m=(s+t)>>1;
	if(l<=m)update(s,m,k<<1,l,r,c);
	if(m<r)update(m+1,t,k<<1|1,l,r,c);
	pushup(k);
}
long long query(int s,int t,int k,int l,int r)
{
	if(l<=s&&t<=r)return sum[k];
	pushdown(k,t-s+1);
	int m=(s+t)>>1;
	long long res=0;
	if(l<=m)res+=query(s,m,k<<1,l,r);
	if(m<r)res+=query(m+1,t,k<<1|1,l,r);
	return res;
}
int main()
{
	scanf("%d%d",&n,&q);
	build(1,n,1);
	for(int i=1;i<=q;i++){
		char ak[3];
		scanf("%s",ak);
		if(ak[0]=='C'){
			int a,b,c;
			scanf("%d%d%d",&a,&b,&c);
			update(1,n,1,a,b,c);
		}
		else{
			int a,b;
			scanf("%d%d",&a,&b);
			printf("%lld\n",query(1,n,1,a,b));
		}
	}
	return 0;
}