CodeForces 315B(线段树+区间更新)

最新推荐文章于 2021-09-27 19:49:24 发布

wust_cyl

最新推荐文章于 2021-09-27 19:49:24 发布

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分类专栏：数据结构文章标签： CodeForces 315B

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本文介绍了一种使用线段树解决区间更新问题的方法，并通过一个具体的编程实例详细展示了线段树的构建、下推、更新及查询操作。适用于需要高效处理数组区间更新与查询的应用场景。

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问题描述:

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

Make vi-th array element equal to xi. In other words, perform the assignment avi = xi.
Increase each array element by yi. In other words, perform n assignments ai = ai + yi (1 ≤ i ≤ n).
Take a piece of paper and write out the qi-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, ..., an (1 ≤ ai ≤ 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 ti (1 ≤ ti ≤ 3) that represents the operation type. If ti = 1, then it is followed by two integers vi and xi, (1 ≤ vi ≤ n, 1 ≤ xi ≤ 109). If ti = 2, then it is followed by integer yi (1 ≤ yi ≤ 104). And if ti = 3, then it is followed by integer qi (1 ≤ qi ≤ 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

问题分析:线段树的区间更新!

代码如下:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;

const int maxn=1e5+100;
struct note
{
    int l,r,val;
}tree[maxn*4];
int a[maxn];

void build(int root,int ll,int rr)
{
    tree[root].l=ll;
    tree[root].r=rr;
    tree[root].val=0;
    if (ll==rr) {
        tree[root].val=a[ll];
        return ;
    }
    int mid=(tree[root].l+tree[root].r)>>1;
    build(root<<1,ll,mid);
    build(root<<1|1,mid+1,rr);
}
void pushdown(int root)
{
    if (tree[root].val) {
        tree[root<<1].val+=tree[root].val;
        tree[root<<1|1].val+=tree[root].val;
        tree[root].val=0;
    }
}
void update(int root,int l,int r,int cur,int p)
{
    int ll=tree[root].l;
    int rr=tree[root].r;
    if (l<=ll&&r>=rr) {
        if (p==2) tree[root].val+=cur;
        else tree[root].val=cur;
        return ;
    }
    pushdown(root);
    int mid=(ll+rr)>>1;
    if (l<=mid) update(root<<1,l,r,cur,p);
    if (r>mid) update(root<<1|1,l,r,cur,p);
}
int query(int root,int pos)
{
    int l=tree[root].l;
    int r=tree[root].r;
    if (l==r) {
        return tree[root].val;
    }
    pushdown(root);
    int mid=(l+r)>>1;
    if (pos<=mid)
        return query(root<<1,pos);
    else
        return query(root<<1|1,pos);
}
int main()
{
    int n,m;
    while (scanf("%d%d",&n,&m)!=EOF) {
        for (int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        build(1,1,n);
        for (int i=1;i<=m;i++) {
            int t;
            scanf("%d",&t);
            if (t==1) {
                int pos,x;
                scanf("%d%d",&pos,&x);
                update(1,pos,pos,x,1);
            }
            else if (t==2) {
                int v;
                scanf("%d",&v);
                update(1,1,n,v,2);
            }
            else if (t==3) {
                int pos;
                scanf("%d",&pos);
                printf("%d\n",query(1,pos));
            }
        }
    }
    return 0;
}