【POJ3468】【树状数组区间修改】A Simple Problem with Integers

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.

Source

【分析】

以前一直以为树状数组只能单点修改，区间询问，今天算是见识到了树状数组的区间修改了。

不说多的了，很巧妙的感觉，对于一个修改操作(a,b,c)，将[a,b]区间整体加上一个c

可以把它拆开成两个部分(a,n,c),和(b + 1,n,-c)，这个时候就好做了，用两个树状数组。

一个树状数组记录从修改点开始一直到结束所增加的所有量，及c*(n-a+1)。另一个记录下增量值c。

那么对于一个在a点右边而在b点左边的点x,它所得到的增量为c*(n-a+1) - c*(n-x)。很好维护了。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <utility>
#include <iomanip>
#include <string>
#include <cmath>
#include <map>

const int MAXN = 100000 + 10; 
const int MAX = 32000 + 10; 
using namespace std;
typedef long long ll;
int n, m;
ll sum[MAXN];//记录原始序列
ll C[2][MAXN]; 

int lowbit(int x){return x&-x;}
ll get(int k, int x){
    ll cnt = 0;
    while (x > 0){
          cnt += C[k][x];
          x -= lowbit(x);
    }
    return cnt;
}
void add(int k, int x, ll val){
     while (x <= n){
           C[k][x] += val;
           x += lowbit(x); 
     }
     return ;
}
void work(){
     memset(C, 0, sizeof(C));//0记录总值、1录增量 
     sum[0] = 0;
     scanf("%d%d", &n, &m);
     for (int i = 1; i <= n; i++){
         scanf("%lld", &sum[i]);
         sum[i] += sum[i - 1];
     }
     for (int i = 1; i <= m; i++){
         char str[2];
         scanf("%s", str);
         if (str[0] == 'Q'){
            int l, r;
            scanf("%d%d", &l, &r);
            printf("%lld
", sum[r] - sum[l - 1] +(get(0, r) - get(1, r) * (n - r)) - (get(0, l - 1) - get(1, l - 1) * (n - l + 1)));
         }else{
               int l, r;
               ll x;
               scanf("%d%d%lld", &l, &r, &x);
               add(0, l, (n - l + 1) * x);
               add(0, r + 1, (n - r) * (-x));
               add(1, l, x);
               add(1, r + 1, -x); 
         }
     }
     //printf("
%d
", get(0, 10));
}

int main(){
    int T;
    #ifdef LOCAL
    freopen("data.txt",  "r",  stdin);
    freopen("out.txt",  "w",  stdout); 
    #endif
    work();
    return 0;
}