Time limit 4000 ms Memory limit 262144 kB

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:

Print operation l, r. Picks should write down the value of . Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r). 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.

題目分析

此題關鍵在於取膜的一個性質 一個數$x$被取膜到0的最多隻要$logx$步 顯然為了讓每次$x$被取膜後的值都儘量大，每一次取膜的值都為$x/2$ 相當於每次除以2，所以最多取$logx$次膜即可到0

於是我們可以用線段樹維護區間和的同時在維護一個區間最大值 每次區間修改進入左右區間前先檢查其區間最大值是否小於取膜值，若是則不進入修改這個區間 對於其他的直接修改到最低端

#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long lt;

lt read()
{
    lt f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}
    while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}
    return f*x;
}

const int maxn=100010;
int n,m;
lt a[maxn];
lt sum[maxn<<2],mx[maxn<<2];

void pushup(int p)
{
	sum[p]=sum[p<<1]+sum[p<<1|1];
	mx[p]=max(mx[p<<1],mx[p<<1|1]); 
}

void build(int s,int t,int p)
{
	if(s==t){ sum[p]=mx[p]=a[s]; return;}
	int mid=s+t>>1;
	build(s,mid,p<<1);build(mid+1,t,p<<1|1);
	pushup(p);
}

void update(int ll,int rr,int s,int t,int p,lt x)
{
	if(s==t){ sum[p]%=x; mx[p]%=x; return;}
	int mid=s+t>>1;
	if(ll<=mid&&mx[p<<1]>=x) update(ll,rr,s,mid,p<<1,x);
	if(rr>mid&&mx[p<<1|1]>=x) update(ll,rr,mid+1,t,p<<1|1,x);
	pushup(p);
}

lt qsum(int ll,int rr,int s,int t,int p)
{
	if(ll<=s&&t<=rr) return sum[p];
	int mid=s+t>>1;lt ans=0;
	if(ll<=mid) ans+=qsum(ll,rr,s,mid,p<<1);
	if(rr>mid) ans+=qsum(ll,rr,mid+1,t,p<<1|1);
	return ans;
}

void change(int u,int s,int t,int p,lt x)
{
	if(s==t){ sum[p]=mx[p]=x; return;}
	int mid=s+t>>1;
	if(u<=mid) change(u,s,mid,p<<1,x);
	else change(u,mid+1,t,p<<1|1,x);
	pushup(p);
}

int main()
{
    n=read();m=read();
    for(int i=1;i<=n;++i) a[i]=read();
    build(1,n,1);
    
    while(m--)
    {
    	lt opt=read(),ll=read(),rr=read();
    	if(opt==1) printf("%lld\n",qsum(ll,rr,1,n,1));
    	else if(opt==2) update(ll,rr,1,n,1,read());
    	else if(opt==3) change(ll,1,n,1,rr);
	}
	return 0;
}