poj3468树状数组的区间更新，区间求和

                                                                                        A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 47174		Accepted: 13844
Case Time Limit: 2000MS

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 a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000. "Q a b" 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

Hint

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

 

解法：

一个树状数组，需要区间更新，区间求和，本来树状数组是区间更新单点求值或单点更新区间求和的，这个它是根据一些公式实现的：

 首先，看更新操作update(s, t, d)把区间A[s]...A[t]都增加d，我们引入一个数组delta[i]，表示

A[i]...A[n]的共同增量，n是数组的大小。那么update操作可以转化为：

1）令delta[s] = delta[s] + d，表示将A[s]...A[n]同时增加d，但这样A[t+1]...A[n]就多加了d，所以

2）再令delta[t+1] = delta[t+1] - d，表示将A[t+1]...A[n]同时减d

 

   然后来看查询操作query(s, t)，求A[s]...A[t]的区间和，转化为求前缀和，设sum[i] = A[1]+...+A[i]，则

                            A[s]+...+A[t] = sum[t] - sum[s-1]，

那么前缀和sum[x]又如何求呢？它由两部分组成，一是数组的原始和，二是该区间内的累计增量和, 把数组A的原始

值保存在数组org中，并且delta[i]对sum[x]的贡献值为delta[i]*(x+1-i)，那么

                            sum[x] = org[1]+...+org[x] + delta[1]*x + delta[2]*(x-1) + delta[3]*(x-2)+...+delta[x]*1

                                         = org[1]+...+org[x] + segma(delta[i]*(x+1-i))

                                         = segma(org[i]) + (x+1)*segma(delta[i]) - segma(delta[i]*i)，1 <= i <= x

=segma(org[i]-delta[i]*i)+(x+1)*delta[i], i<=1<=x   //by huicpc0207 修改  这里就可以转化为两个个数组

这其实就是三个数组org[i], delta[i]和delta[i]*i的前缀和，org[i]的前缀和保持不变，事先就可以求出来，delta[i]和

delta[i]*i的前缀和是不断变化的，可以用两个树状数组来维护。

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;
#define ll long long int
ll a[510000];//维护delta[]
ll a1[510000];//维护delta[]*i
ll b[510000];//本来的数组和
int n;
int lowbit(int x)
{
    return x&(-x);
}
void update(ll *arry,int x,int d)
{
    while(x<=n)
    {
        arry[x]+=d;
        x+=lowbit(x);
    }
}
ll fun(ll *arry,int x)
{
   ll sum=0;
   while(x>0)
   {
       sum+=arry[x];
       x-=lowbit(x);
   }
   return sum;
}
int main()
{
     //freopen("int.txt","r",stdin);
    int k;
    int x,i,y,z;
    scanf("%d%d",&n,&k);
    memset(b,0,sizeof(b));
    memset(a,0,sizeof(a));
    memset(a1,0,sizeof(a1));
    for(i=1;i<=n;i++)
    {
        scanf("%d",&x);
        b[i]+=b[i-1]+x;
    }
    char c;
    for(i=0;i<k;i++)
    {
        c=getchar();
        c=getchar();
        if(c=='C')
        {
            scanf("%d%d%d",&x,&y,&z);
            update(a,x,z);
            update(a,y+1,-z);
            update(a1,x,z*x);
            update(a1,y+1,-z*(y+1));
        }
        else
        {
            scanf("%d%d",&x,&y);
            ll sum=-b[x-1]-x*fun(a,x-1)+fun(a1,x-1);
            sum+=b[y]+(y+1)*fun(a,y)-fun(a1,y);
            printf("%I64d
",sum);
        }
    }
}