CF438D The Child and Sequence

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ShadyPi

最新推荐文章于 2024-02-07 13:51:21 发布

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原题链接：http://codeforces.com/problemset/problem/438/D

The Child and Sequence

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 $a[1], a[2], \cdots, a[n]$ . 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 $\sum_{i=l}^ra[i]$ .
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: $n, m (1 ≤ n, m ≤ 10^5)$ . The second line contains n integers, separated by space: $a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 10^9)$ — initial value of array elements.

Each of the next m lines begins with a number type $(type\in\{1,2,3\})$ .

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: $l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 10^9)$ , which correspond the operation 2.
If type = $3$ , 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.

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 $5, 1 + 2 + 2 = 5$ .

题解

如果取模生效的话，被%的数至少会变成原来的一半，所以我们统计一下区间最大值，如果 $max\ge mod$ 就暴力递归下去取模，最多取 $log_2max$ 次。因为修改是单点的，所以对复杂度的影响不大，最终复杂度为 $O(n\ log_2n\ log_2max)$ 。

代码

#include<bits/stdc++.h>
#define ll long long
#define ls v<<1
#define rs ls|1
using namespace std;
const int M=4e5+5;
ll que[M],mx[M],sum[M],n,m;
void up(int v){mx[v]=max(mx[ls],mx[rs]),sum[v]=sum[ls]+sum[rs];}
void build(int v,int le,int ri)
{
    if(le==ri){mx[v]=sum[v]=que[le];return;}
    int mid=le+ri>>1;
    build(ls,le,mid);build(rs,mid+1,ri);
    up(v);
}
ll ask(int v,int le,int ri,int lb,int rb)
{
    if(lb<=le&&ri<=rb)return sum[v];
    int mid=le+ri>>1;ll ans=0;
    if(lb<=mid)ans=ask(ls,le,mid,lb,rb);
    if(mid<rb)ans+=ask(rs,mid+1,ri,lb,rb);
    return ans;
}
void mod(int v,int le,int ri,int lb,int rb,int d)
{
    if(le==ri){mx[v]=sum[v]=mx[v]%d;return;}
    int mid=le+ri>>1;
    if(lb<=mid&&mx[ls]>=d)mod(ls,le,mid,lb,rb,d);
    if(mid<rb&&mx[rs]>=d)mod(rs,mid+1,ri,lb,rb,d);
    up(v);
}
void alter(int v,int le,int ri,int pos,int d)
{
    if(le==ri){sum[v]=mx[v]=d;return;}
    int mid=le+ri>>1;
    if(pos<=mid)alter(ls,le,mid,pos,d);
    else alter(rs,mid+1,ri,pos,d);
    up(v);
}
void in(){scanf("%d%d",&n,&m);for(int i=1;i<=n;++i)scanf("%d",&que[i]);}
void ac()
{
    build(1,1,n);int a,b,c,op;
    for(int i=1;i<=m;++i)
    {
        scanf("%d%d%d",&op,&a,&b);
        if(op==1)printf("%I64d\n",ask(1,1,n,a,b));
        else if(op==2)scanf("%d",&c),mod(1,1,n,a,b,c);
        else alter(1,1,n,a,b);
    }
}
int main(){in();ac();}