POJ-3468 A Simple Problem with Integers(线段树)

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

看了某个大神写的线段树，感觉要改习惯了555 因为这个大神的线段树方式比以前我习惯写的内个要方便

这里说一下延迟标记或者说懒惰标记的意思

举个栗子，如果让[1,5]里面所有数加1，那么当二分到的这个区间如果正好包含在[1,5]里面，那么为了节约查询的时间，将add+=1 sum+=1*len，后面的子节点可以不考虑。如果后来查询或者更新这个区间的子节点的话将原来这个标记下传，因为原来由于直接包含了这个区间，为了节约时间就没去更新他的子节点，现在需要更新他的子节点了，所以要标记下传 (⊙o⊙)如果还是不懂，那就给你原本zhx大神给我的建议：自己手跑一遍

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <queue>
 6 #include <vector>
 7 #include <stack>
 8 #include <iostream>
 9 #include "algorithm"
10 using namespace std;
11 typedef long long LL;
12
13 #define lson rt<<1,l,m
14 #define rson rt<<1|1,m+1,r
15
16 const int MAX=100005;
17
18 int n,m;
19 LL sum[MAX<<2];
20 LL add[MAX<<2];
21
22 void PushUp(int rt){sum[rt]=sum[rt<<1]+sum[rt<<1|1];}
23
24 void PushDown(int rt,int m){
25     if (add[rt])
26     {add[rt<<1]+=add[rt];
27      add[rt<<1|1]+=add[rt];
28      sum[rt<<1]+=add[rt]*(LL)(m-(m>>1));//很重要！！一开始被坑了！！原因可以自己举个栗子算一下
29      sum[rt<<1|1]+=add[rt]*(LL)(m>>1);
30      add[rt]=0;
31     }
32 }
33
34 void build(int rt,int l,int r){
35     add[rt]=0;
36     if (l==r)
37     {scanf("%lld",&sum[rt]);
38      return;
39     }
40     int m=(l+r)>>1;
41     build(lson);
42     build(rson);
43     PushUp(rt);
44 }
45
46 void update(int rt,int l,int r,int x,int y,int z){
47     if (x<=l && r<=y)
48     {add[rt]+=(LL)z;
49      sum[rt]+=(LL)z*(LL)(r-l+1);
50      return;
51     }
52     PushDown(rt,r-l+1);
53     int m=(l+r)>>1;
54     if (x<=m)
55      update(lson,x,y,z);
56     if (y>m)
57      update(rson,x,y,z);
58     PushUp(rt);
59 }
60
61 LL search(int rt,int l,int r,int x,int y){
62     if (x<=l && r<=y)
63      return sum[rt];
64     PushDown(rt,r-l+1);
65     int m=(l+r)>>1;
66     LL ans(0);
67     if (x<=m)
68      ans+=search(lson,x,y);
69     if (y>m)
70      ans+=search(rson,x,y);
71     return ans;
72 }
73 int main(){
74     freopen ("intergers.in","r",stdin);
75     freopen ("intergers.out","w",stdout);
76     int i,j;
77     int x,y,z;
78     char c;
79     scanf("%d%d",&n,&m);
80     build(1,1,n);
81     for (i=1;i<=m;i++)
82     {c=getchar();
83      while (!(c==‘Q‘ || c==‘C‘)) c=getchar();
84      if (c==‘Q‘)
85      {scanf("%d%d",&x,&y);
86       printf("%lld\n",search(1,1,n,x,y));
87      }
88      else
89      {scanf("%d%d%d",&x,&y,&z);
90       update(1,1,n,x,y,z);
91      }
92     }
93     return 0;
94 }