CodeForces - 438D The Child and Sequence【线段树】

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.

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题目分析

此题关键在于取膜的一个性质
一个数$x$被取膜到0的最多只要$logx$步
显然为了让每次$x$被取膜后的值都尽量大，每一次取膜的值都为$x/2$
相当于每次除以2，所以最多取$logx$次膜即可到0

于是我们可以用线段树维护区间和的同时在维护一个区间最大值
每次区间修改进入左右区间前先检查其区间最大值是否小于取膜值，若是则不进入修改这个区间
对于其他的我们甚至可以不用打延迟标记，直接修改到最低端

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#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;
}