codeforces 315 B.Sereja and Array(线段树区间更新+单点更新+单点询问)-蒲公英云

codeforces 315 B.Sereja and Array(线段树区间更新+单点更新+单点询问)

Bertha 。 2022-06-06 03:57 297阅读 0赞

Sereja has got an array, consisting of n integers, a1, a2, …, a**n. Sereja is an active boy, so he is now going to complete m operations. Each operation will have one of the three forms:

Make v**i-th array element equal to x**i. In other words, perform the assignment avi = x**i.
Increase each array element by y**i. In other words, perform n assignments a**i = a**i + y**i (1 ≤ i ≤ n).
Take a piece of paper and write out the q**i-th array element. That is, the element aqi.

Help Sereja, complete all his operations.

Input

The first line contains integers n, m (1 ≤ n, m ≤ 105). The second line contains nspace-separated integers a1, a2, …, a**n (1 ≤ a**i ≤ 109) — the original array.

Next m lines describe operations, the i-th line describes the i-th operation. The first number in the i-th line is integer t**i (1 ≤ t**i ≤ 3) that represents the operation type. If t**i = 1, then it is followed by two integers v**i and x**i, (1 ≤ v**i ≤ n, 1 ≤ x**i ≤ 109). If t**i = 2, then it is followed by integer y**i (1 ≤ y**i ≤ 104). And if t**i = 3, then it is followed by integer q**i (1 ≤ q**i ≤ n).

Output

For each third type operation print value aqi. Print the values in the order, in which the corresponding queries follow in the input.

Example

Input

10 11
1 2 3 4 5 6 7 8 9 10
3 2
3 9
2 10
3 1
3 10
1 1 10
2 10
2 10
3 1
3 10
3 9

Output

题解：

题意：

操作1 x y表示将a[x]=y

操作2 x 表示全体数列值加上x

操作3 x 表示询问a[x]处的值

思路：

比较裸的一道线段树的题，然后这里的区间更新因为是全体更新所以可以偷懒

代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
using namespace std;
#define lson k*2
#define rson k*2+1
#define M (t[k].l+t[k].r)/2
struct node
{
    int v;
    int tag;
    int l,r;
}t[100005*4];
void pushdown(int k)
{
    if(t[k].tag)
    {
        t[lson].tag+=t[k].tag;
        t[rson].tag+=t[k].tag;
        if(t[lson].l==t[lson].r)
            t[lson].v+=t[k].tag;
        if(t[rson].l==t[rson].r)
            t[rson].v+=t[k].tag;
        t[k].tag=0;
    }
}
void Build(int l,int r,int k)
{
    t[k].l=l;
    t[k].r=r;
    t[k].tag=0;
    if(l==r)
    {
        scanf("%d",&t[k].v);
        return;
    }
    int mid=M;
    Build(l,mid,lson);
    Build(mid+1,r,rson);
}
void update(int k,int v)
{
    t[k].v+=v;
    t[k].tag+=v;
}
void change(int x,int v,int k)
{
    if(t[k].l==x&&t[k].r==x)
    {
        t[k].v=v;
        return;
    }
    pushdown(k);
    int mid=M;
    if(x<=mid)
        change(x,v,lson);
    else
        change(x,v,rson);
}
int query(int x,int k)
{
    if(t[k].l==x&&t[k].r==x)
    {
        return t[k].v;
    }
    pushdown(k);
    int mid=M;
    if(x<=mid)
        return query(x,lson);
    else
        return query(x,rson);
}
int main()
{
    int i,j,n,m,d,x,y;
    scanf("%d%d",&n,&m);
    Build(1,n,1);
    while(m--)
    {
        scanf("%d",&d);
        if(d==1)
        {
            scanf("%d%d",&x,&y);
            change(x,y,1);
        }
        else if(d==2)
        {
            scanf("%d",&x);
            update(1,x);
        }
        else
        {
            scanf("%d",&x);
            printf("%d\n",query(x,1));
        }
    }
    return 0;
}