【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 =  + ;
 const int MAX =  + ;
 using namespace std;
 typedef long long ll;
 int n, m;
 ll sum[MAXN];//记录原始序列
 ll C[][MAXN]; 
 
 int lowbit(int x){return x&-x;}
 ll get(int k, int x){
     ll cnt = ;
     while (x > ){
           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, , sizeof(C));//0记录总值、1录增量
      sum[] = ;
      scanf("%d%d", &n, &m);
      for (int i = ; i <= n; i++){
          scanf("%lld", &sum[i]);
          sum[i] += sum[i - ];
      }
      for (int i = ; i <= m; i++){
          char str[];
          scanf("%s", str);
          if (str[] == 'Q'){
             int l, r;
             scanf("%d%d", &l, &r);
             printf("%lld\n", sum[r] - sum[l - ] +(get(, r) - get(, r) * (n - r)) - (get(, l - ) - get(, l - ) * (n - l + )));
          }else{
                int l, r;
                ll x;
                scanf("%d%d%lld", &l, &r, &x);
                add(, l, (n - l + ) * x);
                add(, r + , (n - r) * (-x));
                add(, l, x);
                add(, r + , -x);
          }
      }
      //printf("\n%d\n", get(0, 10));
 }
 
 int main(){
     int T;
     #ifdef LOCAL
     freopen("data.txt",  "r",  stdin);
     freopen("out.txt",  "w",  stdout);
     #endif
     work();
     return ;
 }