POJ - 3468A Simple Problem with Integers （线段树区间更新，区间查询和）

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

解题思路：利用懒惰标记更新区间并求区间和

代码如下：

  1 #include<iostream>
  2 #include<stdio.h>
  3 using namespace std;
  4
  5 const int MAXN = 500005;
  6 int a[MAXN];
  7 long long int tree[MAXN*4];
  8 long long int lazy[MAXN*4];
  9 int N,Q;
 10 string s;
 11 int x , y, z;
 12 void push_down(int l ,int r ,int rt)
 13 {
 14     int m = (l+r)/2;
 15
 16     if(lazy[rt])
 17     {
 18         tree[rt*2] += lazy[rt]*(m-l+1);
 19         tree[rt*2+1] += lazy[rt]*(r-m);
 20         lazy[rt*2] += lazy[rt];
 21         lazy[rt*2+1] += lazy[rt];
 22         lazy[rt] = 0;
 23     }
 24 }
 25 void bulid_tree(int l ,int r ,int rt)
 26 {
 27     if(l==r)
 28     {
 29         tree[rt] = a[l];
 30         return ;
 31     }
 32     int m = (l+r)/2;
 33
 34     bulid_tree(l,m,rt*2);
 35     bulid_tree(m+1,r,rt*2+1);
 36     tree[rt] = tree[rt*2]+tree[rt*2+1];
 37 }
 38
 39 long long int Query(int x ,int y ,int l ,int r ,int rt)
 40 {
 41     long long sum = 0 ;
 42     if(x<=l&&r<=y)
 43     {
 44         return tree[rt];
 45     }
 46     int m = (l+r)/2;
 47     push_down(l,r,rt);
 48     if(x<=m)
 49     {
 50         sum += Query(x,y,l,m,rt*2);
 51     }
 52     if(m<y)
 53     {
 54         sum += Query(x,y,m+1,r,rt*2+1);
 55     }
 56     return sum;
 57 }
 58 void Update(int x ,int y ,int k ,int l ,int r ,int rt)
 59 {
 60     if(x<=l&&y>=r)
 61     {
 62         tree[rt] += k*(r-l+1);
 63         lazy[rt] += k;
 64         return ;
 65     }
 66     push_down(l,r,rt);
 67     int m = (l+r)/2;
 68     if(x<=m)
 69     {
 70         Update(x,y,k,l,m,rt*2);
 71     }
 72     if(y>m)
 73     {
 74         Update(x,y,k,m+1,r,rt*2+1);
 75     }
 76     tree[rt] = tree[rt*2]+tree[rt*2+1];
 77 }
 78 int main()
 79 {
 80     scanf("%d%d",&N,&Q);
 81     for(int i = 1 ; i <= N;i++)
 82     {
 83         scanf("%d",&a[i]);
 84     }
 85     bulid_tree(1,N,1);
 86     while(Q--)
 87     {
 88         cin>>s;
 89         if(s[0]==‘Q‘)
 90         {
 91             scanf("%d%d",&x,&y);
 92             printf("%lld\n",Query(x,y,1,N,1));
 93
 94         }
 95         else
 96         if(s[0]==‘C‘)
 97         {
 98             scanf("%d%d%d",&x,&y,&z);
 99             Update(x,y,z,1,N,1);
100         }
101     }
102     return 0;
103 }

原文地址：https://www.cnblogs.com/yewanting/p/10800242.html