Sereja and Array-数组操作或者线段树或树状数组

CodeForces - 315B

Sereja and Array

Time Limit: 1000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

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:

1. Make v_i-th array element equal to x_i. In other words,
   perform the assignment a_vi = x_i.
2. Increase each array element by y_i. In other words, perform n assignments a_i = a_i + y_i(1 ≤ i ≤ n).
3. Take a piece of paper and write out the q_i-th array element. That is, the element a_qi.

Help Sereja, complete all his operations.

Input

The first line contains integers n, m(1 ≤ n, m ≤ 10⁵).
The second line contains n space-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 a_qi. Print the values in the order, in which the corresponding queries
follow in the input.

Sample Input

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

这道题目大家须要思考，不要一看到题目就用线段树，要想想有没有更好的方法，这里的题目给出的三种操作

能够知道没有一个是对区间进行操作的，唯一一个都是对整个数组操作。对全部的数的影响一样。

所以代码便能够变为例如以下：

/*
Author: 2486
Memory: 204 KB		Time: 93 MS
Language: GNU G++11 4.9.2		Result: Accepted
VJ RunId: 4206974		Real RunId: 12270208
Public:		No Yes
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=1e5+5;
int n,m,p,c,v,a[maxn];
int main() {
    scanf("%d%d",&n,&m);
    for(int i=1; i<=n; i++) {
        scanf("%d",&a[i]);
    }
    int cnt=0;
    while(m--) {
        scanf("%d%d",&p,&c);
        if(p==1) {
            scanf("%d",&v);
            a[c]=v-cnt;
        } else if(p==2) {
            cnt+=c;
        } else {
            printf("%d\n",a[c]+cnt);
        }
    }
    return 0;
}