线段树【CF620E】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 mm 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,mn,m (1<=n,m<=10^{5})(1<=n,m<=105) . The second line contains nn integers, separated by space: $ a[1],a[2],...,a[n] (1<=a[i]<=10^{9}) $ — initial value of array elements.

Each of the next mm lines begins with a number typetype .

• If type=1type=1 , there will be two integers more in the line: $ l,r (1<=l<=r<=n) $ , which correspond the operation 1.
• If type=2type=2 , there will be three integers more in the line: $ l,r,x (1<=l<=r<=n; 1<=x<=10^{9}) $ , which correspond the operation 2.
• If type=3type=3 , there will be two integers more in the line: $ k,x (1<=k<=n; 1<=x<=10^{9}) $ , 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.

题目大意：

• 给出一个序列，进行如下三种操作：
• 区间求和
• 区间每个数模 xx
• 单点修改
• n,m≤100000

裸的线段树问题.,但是问题在于如何取模。

很容易想到的是,如果区间的最大值比取模的数小,那么我们就不需要修改。

因此，我们维护区间最大值。

但是如何修改？我们需要知道其位置。

因此,我们维护最大值位置,然后单点修改即可。

每次判断区间最大值时候比取模的数小。

如果小,那我们就不用取模,所以就可以切掉这个题了！

代码

#include<cstdio>
#include<iostream>
#include<algorithm>
#define int long long
#define R register

using namespace std;

const int gz=1e5+8;

inline void in(int &x)
{
	int f=1;x=0;char s=getchar();
	while(!isdigit(s)){if(s=='-')f=-1;s=getchar();}
	while(isdigit(s)){x=x*10+s-'0';s=getchar();}
	x*=f;
}

#define ls o<<1
#define rs o<<1|1

int tr[gz<<2],mx[gz<<2],val[gz],n,m;

inline int idmax(R int x,R int y)
{
	return val[x]>val[y] ? x:y;
}

inline void up(R int o)
{
	mx[o]=idmax(mx[ls],mx[rs]);
	tr[o]=tr[ls]+tr[rs];
}

void build(R int o,R int l,R int r)
{
	if(l==r)
	{
		tr[o]=val[l];
		mx[o]=l;
		return;
	}
	R int mid=(l+r)>>1;
	build(ls,l,mid);
	build(rs,mid+1,r);
	up(o);
}

void change(R int o,R int l,R int r,R int pos,R int del)
{
	if(l==r){tr[o]=val[l];return;}
	R int mid=(l+r)>>1;
	if(pos<=mid)change(ls,l,mid,pos,del);
	else change(rs,mid+1,r,pos,del);
	up(o);
}

int query(R int o,R int l,R int r,R int x,R int y)
{
	if(x<=l and y>=r)return tr[o];
	R int mid=(l+r)>>1,res=0;
	if(x<=mid)res+=query(ls,l,mid,x,y);
	if(y>mid)res+=query(rs,mid+1,r,x,y);
	return res;
}

int query_max(R int o,R int l,R int r,R int x,R int y)
{
	if(l==x and y==r) return mx[o];
	R int mid=(l+r)>>1;
	if(y<=mid) return query_max(ls,l,mid,x,y);
	else if(x>mid)return query_max(rs,mid+1,r,x,y);
	else return idmax(query_max(ls,l,mid,x,mid),query_max(rs,mid+1,r,mid+1,y));
}

signed main()
{
	in(n);in(m);
	for(R int i=1;i<=n;i++)in(val[i]);
	build(1,1,n);
	for(R int l,r,k,opt;m;m--)
	{
		in(opt);
		switch(opt)
		{
			case 1:in(l),in(r),printf("%lld\n",query(1,1,n,l,r));break;
			case 2:break;
			case 3:in(l),in(r);val[l]=r;change(1,1,n,l,r);break;
		}
		if(opt==2)
		{
			in(l),in(r),in(k);
			for(R int pos;;)
			{
				pos=query_max(1,1,n,l,r);
				if(val[pos]<k)break;
				val[pos]%=k;
				change(1,1,n,pos,val[pos]);
			}
		}
	}
}