CF 315B(Sereja and Array-峰顶距离统计)

B. Sereja and Array

time limit per test

memory limit per test

input

output

n integers, a₁, a₂, ..., a_n. Sereja is an active boy, so he is now going to complete m

1. v_i-th array element equal tox_i. In other words, perform the assignmenta_vi=x_i.
2. y_i. In other words, performnassignmentsa_i=a_i+y_i(1 ≤i≤n).
3. q_i-th array element. That is, the elementa_qi.

Help Sereja, complete all his operations.

Input

n, m (1 ≤ n, m ≤ 10⁵). The second line contains n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹)

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

a_qi. Print the values in the order, in which the corresponding queries follow in the input.

Sample test(s)

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

2
9
11
20
30
40
39

水题，统计与峰顶的距离，模拟

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define MAXM (1000000+10)
int n,m;
/*
struct tree_arr
{
    int a[MAXM*4];
    tree_arr(){memset(a,0,sizeof(a));}
    void inc(int x,int c)
    {
        for(int i=x;i<=n;i+=i&(-i)) a[x]+=c;
    }
    int qur(int x)
    {
        int ans=0;
        for(int i=x;i;i-=i&(-i)) ans+=a[x];
        return ans;
    }
}T1,T2;
void Ins()
{

}*/
int a[MAXM],tmp=0;

int main()
{
//  freopen(".in","r",stdin);
//  freopen(".out","w",stdout);
    cin>>n>>m;
    For(i,n) {scanf("%d",&a[i]);}
    For(i,m)
    {
        int type;
        scanf("%d",&type);
        if (type==1)
        {
            int v,x;
            scanf("%d%d",&v,&x);
            a[v]+=x-(a[v]+tmp);
        }
        else if (type==2)
        {
            int w;
            cin>>w;
            tmp+=w;
        }
        else
        {
            int v;
            scanf("%d",&v);
            cout<<a[v]+tmp<<endl;
        }

    }

    return 0;
}