A Simple Problem with Integers||POJ3468

link:http://poj.org/problem?id=3468
Description

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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

树状数组天生用来动态维护数组前缀和，其特点是每次更新一个元素的值，查询只能查数组的前缀和，但这个题目求的是某一区间的数组和，而且要支持批量更新某一区间内元素的值，怎么办呢？实际上，还是可以把问题转化为求数组的前缀和。

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：
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
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。

这是区间和的模板题
AC代码：

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>#include<iostream>using namespace std;long long a1[100010],a2[100010],a3[100010];int n;void update(long long *a,int x,int y){    while(x<=n)    {        a[x]+=y;        x+=(x&-x);    }}long long query(long long *a,int x){    long long sum=0;    while(x>0)    {        sum+=a[x];        x-=(x&-x);    }    return sum;}int main(){    int i,m,a,b,c,x;    char s;    long long ans;     memset(a3,0,sizeof(a3));    scanf("%d%d",&n,&m);    for(i=1;i<=n;i++)    {        scanf("%d",&x);        a3[i]=a3[i-1]+x;    }    while(m--)    {        scanf("%s",&s);        if(s=='Q')        {            scanf("%d%d",&a,&b);            ans=a3[b]-a3[a-1];            ans+=(b+1)*query(a1,b)-query(a2,b);            ans-=a*query(a1,a-1)-query(a2,a-1);            printf("%lld\n",ans);        }        else        {            scanf("%d%d%d",&a,&b,&c);            update(a1,a,c);            update(a1,b+1,-c);            update(a2,a,a*c);            update(a2,b+1,-c*(b+1));        }    }return 0;}