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Sereja and Array-数组操作或者线段树或树状数组

2024-09-20 23:13:31   215人阅读

CodeForces - 315B
Sereja and Array
Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u

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Description

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:

  1. Make vi-th array element equal to xi. In other words, perform the assignment avi?=?xi.
  2. Increase each array element by yi. In other words, perform n assignments ai?=?ai?+?yi(1?≤?i?≤?n).
  3. 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 n space-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.

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;
}















   

Sereja and Array-数组操作或者线段树或树状数组






























        

            
                
                  
                
                

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