线段树之区间更新The Child and Sequence

Description

At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.

Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:

1. Print operation l, r. Picks should write down the value of .
2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i(l ≤ i ≤ r).
3. Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x).

Can you help Picks to perform the whole sequence of operations?

Input

The first line of input contains two integer: n, m(1 ≤ n, m ≤ 10⁵). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 10⁹) — initial value of array elements.

Each of the next m lines begins with a number type.

• If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1.
• If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 10⁹), which correspond the operation 2.
• If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 10⁹), which correspond the operation 3.

Output

For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.

Sample Input

Input

5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3

Output

8
5

Input

10 10
6 9 6 7 6 1 10 10 9 5
1 3 9
2 7 10 9
2 5 10 8
1 4 7
3 3 7
2 7 9 9
1 2 4
1 6 6
1 5 9
3 1 10

Output

49
15
23
1
9

Hint

Consider the first testcase:

• At first, a = {1, 2, 3, 4, 5}.
• After operation 1, a = {1, 2, 3, 0, 1}.
• After operation 2, a = {1, 2, 5, 0, 1}.
• At operation 3, 2 + 5 + 0 + 1 = 8.
• After operation 4, a = {1, 2, 2, 0, 1}.
• At operation 5, 1 + 2 + 2 = 5.

  #include<cstdio>
  #include<iostream>
  using namespace std;
  #define ll long long
  
  const int N=1e5+5;
  ll f[N<<2];//因为是存区间和，所以要开longlong
  ll lazy[N<<2];//存最值不用longlong
  int n,m;
  
  void pushup(int root){
      int rt=root<<1; 
      
      f[root]=f[rt]+f[rt+1];
      lazy[root]=max(lazy[rt],lazy[rt+1]);
  }
  
  
  void build(int left,int right,int root){
      if(left==right){
          
          scanf("%lld",&f[root]);
          lazy[root]=f[root];
          
          return ;
      }
      
      int mid=(left+right)>>1;
      int rt=root<<1;
      
      build(left,mid,rt);
      build(mid+1,right,rt+1);
      
      pushup(root);
  }
  
  void change(int x,int val,int left,int right,int root){//单点修改
      if(left==right){
          lazy[root]=f[root]=val;
          return ;
      }
      
      int mid=(left+right)>>1;
      int rt=root<<1;
      
      if(x<=mid){
          change(x,val,left,mid,rt);
      }else{
          change(x,val,mid+1,right,rt+1);
      }
      
      pushup(root);
  }
  
  void update(int x,int uleft,int uright,int left,int right,int root){//区间取模
      
      if(lazy[root]<x)
      {
          return ;
      }//区间最值如果小于x，则不用继续递归
      
      if(left==right){
          lazy[root]=(f[root]%=x);
          return ;
      }
      
      int rt=root<<1;
      int mid=(left+right)>>1;
      
      if(uleft<=mid){
          update(x,uleft,uright,left,mid,rt);
      }
      
      if(uright>mid){
          update(x,uleft,uright,mid+1,right,rt+1);
      }
      
      pushup(root);
  }
  
  ll query(int qleft,int qright,int left,int right,int root){//区间求和
      if(qleft<=left&&qright>=right){
          return f[root];
      }
      
      int mid=(left+right)/2;
      int rt=root<<1;
      
      ll ans=0;
      
      if(qleft<=mid){
          ans+=query(qleft,qright,left,mid,rt);
      }
      
      if(qright>mid){
          ans+=query(qleft,qright,mid+1,right,rt+1);
      }
      return ans;
  }
  
  int main(){
      
      scanf("%d%d",&n,&m);
      build(1,n,1);
      long long ans=0;
      int op,l,r,x,val;
      while(m--){
          scanf("%d",&op);
          
          if(op==1)
          {
              
              scanf("%d%d",&l,&r);
              ans=query(l,r,1,n,1);
              printf("%lld\n",ans);
              
          }else if(op==2)
          {
          
              scanf("%d%d%d",&l,&r,&x);
              update(x,l,r,1,n,1);
          }
          else
          {
              
              scanf("%d%d",&x,&val);
              change(x,val,1,n,1);
          }
      }
      return 0;
  }

   

标签：10,Sequence,int,线段,Picks,Child,operation,should,line
来源： https://www.cnblogs.com/killjoyskr/p/16492268.html