树状数组成段更新——POJ 3468

A Simple Problem with IntegersCrawling in process...
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Time Limit:5000MS    
Memory Limit:131072KB    
64bit IO Format:%I64d & %I64u

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Status Practice POJ 3468

Description

You have N integers, A₁,
A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤
A_i ≤ 1000000000.

Each of the next Q lines represents an operation.

"C abc" means adding c to each of A_a,
A_a+1, ... ,
A_b. -10000 ≤ c ≤ 10000.

"Q ab" means querying the sum of A_a,
A_a+1, ... ,
A_b.

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

4
55
9
15

这里讲得很详细：点击打开链接

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
const int MAXN=100000+10;
using namespace std;
typedef long long LL;
LL sum[MAXN];
LL c1[MAXN];
LL c2[MAXN];

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

void update(LL *c, int x, LL val)
{
	while(x<=MAXN)
	{
		c[x]+=val;
		x+=lowbit(x);
	}
}

LL Sum(LL *c, int x)
{
	LL res=0;
	while(x>0)
	{
		res+=c[x];
		x-=lowbit(x);
	}
	return res;
}

LL SUM(int x)
{
	return sum[x]+(x+1)*Sum(c1,x)-Sum(c2,x);
}

int main()
{
	//reopen("in.txt","r",stdin);
	int n,m;
	while(scanf("%d%d", &n,&m)==2)
	{
		memset(c1,0,sizeof(c1));
		memset(c2,0,sizeof(c2));
		int i,j,k;
		for(i=1; i<=n; i++){
			scanf("%d", &j);
			sum[i]=sum[i-1]+j;
		}
		char ch[3];
		while(m--)
		{
			scanf("%s", ch);
			if(ch[0]=='Q'){
				scanf("%d%d", &i,&j);
				LL ans=SUM(j)-SUM(i-1);
				printf("%I64d\n", ans);
			}
			else{
				scanf("%d%d%d", &i,&j,&k);
				update(c1,i,k);
				update(c1,j+1,-k);
				update(c2,i,i*k);
				update(c2,j+1,-k*(j+1));
			}
		}
	}
	return 0;
}

树状数组成段更新——POJ 3468