CodeForces - 438D: The Child and Sequence(势能线段树)

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 CodeForces - 438D: The Child and Sequence(势能线段树).
  2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[imod 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 CodeForces - 438D: The Child and Sequence(势能线段树).

  • 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.

Examples

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

Note

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.

题意:给出数组,有三种操作,分别是区间求和,区间取模 ,单点修改。

思路:一个点被取模,那么其大小减半,所以一个数最多被操作log次,这样的话就不难想到势能线段树。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=;
int Mx[maxn<<]; ll sum[maxn<<];
void build(int Now,int L,int R)
{
if(L==R){
scanf("%d",&Mx[Now]);
sum[Now]=Mx[Now]; return ;
}
int Mid=(L+R)>>;
build(Now<<,L,Mid); build(Now<<|,Mid+,R);
Mx[Now]=max(Mx[Now<<],Mx[Now<<|]);
sum[Now]=sum[Now<<]+sum[Now<<|];
}
ll query(int Now,int L,int R,int l,int r){
if(l<=L&&r>=R) return sum[Now];
int Mid=(L+R)>>; ll res=;
if(l<=Mid) res+=query(Now<<,L,Mid,l,r);
if(r>Mid) res+=query(Now<<|,Mid+,R,l,r);
return res;
}
void change(int Now,int L,int R,int pos,int val)
{
if(L==R){
Mx[Now]=val; sum[Now]=val; return ;
}
int Mid=(L+R)>>;
if(pos<=Mid) change(Now<<,L,Mid,pos,val);
else change(Now<<|,Mid+,R,pos,val);
Mx[Now]=max(Mx[Now<<],Mx[Now<<|]);
sum[Now]=sum[Now<<]+sum[Now<<|];
}
void modp(int Now,int L,int R,int l,int r,int P)
{
if(Mx[Now]<P) return ;
if(L==R) {
Mx[Now]%=P; sum[Now]=Mx[Now]; return ;
}
int Mid=(L+R)>>;
if(l<=Mid) modp(Now<<,L,Mid,l,r,P);
if(r>Mid) modp(Now<<|,Mid+,R,l,r,P);
Mx[Now]=max(Mx[Now<<],Mx[Now<<|]);
sum[Now]=sum[Now<<]+sum[Now<<|];
}
int main()
{
int N,M,opt,L,R,P;
scanf("%d%d",&N,&M);
build(,,N);
while(M--){
scanf("%d",&opt);
if(opt==) {
scanf("%d%d",&L,&R);
printf("%I64d\n",query(,,N,L,R));
}
else if(opt==){
scanf("%d%d%d",&L,&R,&P);
modp(,,N,L,R,P);
}
else {
scanf("%d%d",&L,&R);
change(,,N,L,R);
}
}
return ;
}
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