Codeforces Round #250 (Div. 1) D. The Child and Sequence 线段树

D. The Child and Sequence

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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：给你一个序列；

　　　操作1，给你一个区间[l,r]求区间和；

　　   操作2，给你区间每个点取模；

　　　操作3，单点修改；

思路：开始写了个延迟标记，发现那个每个点取模和并不等于区间和取模，例如： 1 2 模3 ，等于3，并不等于0；

　　　更新到每个点显然超时，所以需要一点优化，记录一个最大值，判断最大值是否大于模，不大于则不用更新；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=1e5+10,M=4e6+10,inf=1e9+10;
ll sum[N<<2];
ll maxx[N<<2];
void pushup(int pos)
{
    sum[pos]=sum[pos<<1|1]+sum[pos<<1];
    maxx[pos]=max(maxx[pos<<1|1],maxx[pos<<1]);
}
void buildtree(int l,int r,int pos)
{
    if(l==r)
    {
        scanf("%lld",&sum[pos]);
        maxx[pos]=sum[pos];
        return;
    }
    int mid=(l+r)>>1;
    buildtree(l,mid,pos<<1);
    buildtree(mid+1,r,pos<<1|1);
    pushup(pos);
}
void update3(int point,ll change,int l,int r,int pos)
{
    if(l==r&&l==point)
    {
        sum[pos]=change;
        maxx[pos]=change;
        return;
    }
    int mid=(l+r)>>1;
    if(point<=mid)
    update3(point,change,l,mid,pos<<1);
    else
    update3(point,change,mid+1,r,pos<<1|1);
    pushup(pos);
}
void update2(int L,int R,int m,int l,int r,int pos)
{
    if(l==r)
    {
        sum[pos]%=m;
        maxx[pos]=sum[pos];
        return;
    }
    int mid=(l+r)>>1;
    if(L<=mid&&maxx[pos<<1]>=m)
    update2(L,R,m,l,mid,pos<<1);
    if(R>mid&&maxx[pos<<1|1]>=m)
    update2(L,R,m,mid+1,r,pos<<1|1);
    pushup(pos);
}
ll query(int L,int R,int l,int r,int pos)
{
    if(L<=l&&r<=R)
    return sum[pos];
    int mid=(l+r)>>1;
    ll ans=0;
    if(L<=mid)
    ans+=query(L,R,l,mid,pos<<1);
    if(R>mid)
    ans+=query(L,R,mid+1,r,pos<<1|1);
    return ans;
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    buildtree(1,n,1);
    while(m--)
    {
        int flag,l,r,mo;
        ll x;
        scanf("%d",&flag);
        if(flag==1)
        {
            scanf("%d%d",&l,&r);
            printf("%lld\n",query(l,r,1,n,1));
        }
        else if(flag==2)
        {
            scanf("%d%d%d",&l,&r,&mo);
            update2(l,r,mo,1,n,1);
        }
        else
        {
            scanf("%d%lld",&mo,&x);
            update3(mo,x,1,n,1);
        }
    }
    return 0;
}