A Simple Problem with Integers （树状数组区间修改，区间查询）

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
int t,n,jishu,m;
const int N=100005;
ll cnt1[N],arr[N],cnt2[N];
int lowbits(int x) { return x&-x; }
void add(int x,int value){
    ll temp=x;
    while(x<=N){
        cnt1[x]+=value;
        cnt2[x]+=value*(temp-1);
        x+=lowbits(x);
    }
}
ll query(int x){
    ll sum=0;
    int temp=x;
    while(x>0){
        sum+=temp*cnt1[x]-cnt2[x];
        x-=lowbits(x);
    }
    return sum;
}

int main(){

    scanf("%d%d",&n,&m);
    for(int i=1,d;i<=n;i++) scanf("%lld",&arr[i]),add(i,arr[i]-arr[i-1]);
    char ss[10];
    for(int i=1;i<=m;i++){
        int d1,d2,d3;
        scanf("%s",ss);
        if(ss[0]=='Q'){
            scanf("%d%d",&d1,&d2);
            printf("%lld
",query(d2)-query(d1-1));
        }
        else{
            scanf("%d%d%d",&d1,&d2,&d3);
            add(d1,d3),add(d2+1,-d3);
        }
    }
    return 0;
}