POJ 3468 A Simple Problem with Integers（线段树区间修改）

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.

两种操作，

1.一段连续区间都加上一个数。

2.查询一段区间的数字之和。

第一次写区间修改，套错了模板，各种错误层出不穷，关键还是对线段树的理解不够深刻吧。

#include<iostream>
#include<cstdio>
#define ll long long
#include<cstring>
const int MX = 1e5+5;
ll sum[4 * MX];
ll la[4 * MX];
void pushdown(int rt, int l, int r){
  if(la[rt]){
    int m = (l + r) / 2;
    la[2 * rt] +=  la[rt];
    la[2 * rt + 1] +=  la[rt];            //两端更新 
    sum[2 * rt] += 1LL * (m - l + 1) * la[rt];      //是+= 
    sum[2 * rt + 1] += 1LL * (r - m) * la[rt];
    la[rt] = 0;
  }
}
void build(int l, int r, int rt){
  if(l == r) {
  scanf("%I64d",&sum[rt]);
  return;
  }
  int m = (l + r) / 2;
  build(l, m, 2 * rt);
  build(m+1, r, 2 * rt +1);
  sum[rt] = sum[2 * rt] + sum[2 * rt + 1];
} 
void update (int st, int en, int co, int l, int r, int rt){
  if(st <= l && r <= en){
    sum[rt] += 1LL* (r - l + 1) * co;    //这里错写成 l-r
    la [rt] += co;              //+= 错写 = 
    return ;
  }
  pushdown(rt, l, r);
  int m = (l + r) / 2; 
  if(st <= m) update(st, en, co, l, m, 2 * rt);
  if(m < en)  update(st, en, co, m+1, r, 2 * rt + 1);
  sum[rt] = sum[2 * rt] + sum[2 * rt + 1];
}

ll sea(int st, int en, int l, int r, int rt){
  if(st <= l && r <= en) return sum[rt];
  int m = (l + r) /2 ;
  ll ans = 0;
  pushdown(rt, l, r);            //pushdown不能省 
  if(st <= m) ans += sea(st, en, l, m, 2 * rt);
  if(m < en) ans += sea(st, en, m+1, r, 2 * rt + 1);
  return ans;
}


int main(){
  int n,q,a,b,c;
  char cmd[3];
  while(scanf("%d%d",&n,&q) != EOF){
    build(1, n, 1);
    memset(la, 0, sizeof(la));          //差点sizeof(0);
    while(q--){
      scanf("%s",cmd);
      if(cmd[0] == 'Q'){
        scanf("%d%d",&a,&b);
        printf("%I64d\n",sea(a, b, 1, n, 1));
      }
      else {
      scanf("%d%d%d",&a, &b, &c);
      update(a, b, c, 1, n, 1);
      }
    }
  }
  return 0;
}