POJ 3468 A Simple Problem with Integers（线段树区间求和）

Posted 2020-11-01 ghzz

传送门：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

 

 

题意：

给定一个数列有n个数以及q个操作

有两种操作：

　　c：给出a,b,c 对a，b区间的数同时加上c

　　q：给出a,b 求a,b区间的和

题解：

如果对每个节点，维护对应区间的和，可以在O（logn）时间内求出任意区间的和，但是不能够高效的实现对一个区间同时加一个值，因为需要对其所有关联节点进行更新

为了保持线段树的高效，对于每个节点维护两个数据：

a 给这个节点对应的区间内同时加上一个值

b在这个节点对应的区间内除去a之外其他值的和

代码：

#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <deque>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <utility>
#include <vector>
#define mem(arr, num) memset(arr, 0, sizeof(arr))
#define _for(i, a, b) for (int i = a; i <= b; i++)
#define __for(i, a, b) for (int i = a; i >= b; i--)
#define IO                     \
  ios::sync_with_stdio(false);   cin.tie(0);                   cout.tie(0);
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f;
const double EPS = 1e-10;
const ll mod = 1000000007LL;
const int N = 1 << 19;
ll data[N],datb[N];
int n,q;
void build(int l, int r, int k) {
  data[k] = datb[k] = 0;
  if(l==r) {
    scanf("%d",&datb[k]);return ;
  }
  build(l,(l+r)/2,k<<1);
  build((l+r)/2+1,r,k<<1|1);
  datb[k] = datb[k<<1] + datb[k<<1|1];
}
void update(int a,int b,int k,int l,int r,int dat) {
  if(a <= l && b >= r) {
      data[k] += dat;
  } else if(a <= r && b >= l) {
      datb[k] += (min(b,r) - max(a,l) + 1) * x;
      update(a,b,k<<1,l,(l+r)/2,dat);
      update(a,b,k<<1|1,(l+r)/2+1,r,dat);
  }
}
ll query(int a,int b,int k,int l,int r) {
    if(b < l || a > r) return 0;
    else if(a <= l && b >= r) return data[k] * (r-l+1) + datb[k];
    else {
        ll res = (min(b,r) - max(a,l) + 1) *data[k];
        res += query(a,b,k<<1,l,(l+r)/2);
        res += query(a,b,k<<1|1,(l+r)/2+1);
        return res;
    }
}
int main() {
  scanf("%d%d",&n,&q);
  build(1,n,1);
  char op;
  int a,b,c;
  _for(i, 1, q) {
      getchar();
      scanf("%c",&op);
      if(op == ‘Q‘) {
          scanf("%d%d",&a,&b);
          printf("%lld\n",query(a,b,1,1,n));
      } else {
          scanf("%d%d%d",&a,&b,&c);
          update(a,b,1,1,n,c);
      }
  }
  return 0;
}