A. Adjacent Replacements

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mishka got an integer array aa of length nn as a birthday present (what a surprise!).

Mishka doesn't like this present and wants to change it somehow. He has invented an algorithm and called it "Mishka's Adjacent Replacements Algorithm". This algorithm can be represented as a sequence of steps:

  • Replace each occurrence of 11 in the array aa with 22;
  • Replace each occurrence of 22 in the array aa with 11;
  • Replace each occurrence of 33 in the array aa with 44;
  • Replace each occurrence of 44 in the array aa with 33;
  • Replace each occurrence of 55 in the array aa with 66;
  • Replace each occurrence of 66 in the array aa with 55;
  • ……
  • Replace each occurrence of 109−1109−1 in the array aa with 109109;
  • Replace each occurrence of 109109 in the array aa with 109−1109−1.

Note that the dots in the middle of this algorithm mean that Mishka applies these replacements for each pair of adjacent integers (2i−1,2i2i−1,2i) for each i∈{1,2,…,5⋅108}i∈{1,2,…,5⋅108} as described above.

For example, for the array a=[1,2,4,5,10]a=[1,2,4,5,10], the following sequence of arrays represents the algorithm:

[1,2,4,5,10][1,2,4,5,10] →→ (replace all occurrences of 11 with 22) →→ [2,2,4,5,10][2,2,4,5,10] →→ (replace all occurrences of 22 with 11) →→ [1,1,4,5,10][1,1,4,5,10] →→(replace all occurrences of 33 with 44) →→ [1,1,4,5,10][1,1,4,5,10] →→ (replace all occurrences of 44 with 33) →→ [1,1,3,5,10][1,1,3,5,10] →→ (replace all occurrences of 55 with 66) →→ [1,1,3,6,10][1,1,3,6,10] →→ (replace all occurrences of 66 with 55) →→ [1,1,3,5,10][1,1,3,5,10] →→ …… →→ [1,1,3,5,10][1,1,3,5,10] →→ (replace all occurrences of 1010 with 99) →→ [1,1,3,5,9][1,1,3,5,9]. The later steps of the algorithm do not change the array.

Mishka is very lazy and he doesn't want to apply these changes by himself. But he is very interested in their result. Help him find it.

Input

The first line of the input contains one integer number nn (1≤n≤10001≤n≤1000) — the number of elements in Mishka's birthday present (surprisingly, an array).

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the elements of the array.

Output

Print nn integers — b1,b2,…,bnb1,b2,…,bn, where bibi is the final value of the ii-th element of the array after applying "Mishka's Adjacent Replacements Algorithm" to the array aa. Note that you cannot change the order of elements in the array.

Examples

input

Copy

5
1 2 4 5 10

output

Copy

1 1 3 5 9

input

Copy

10
10000 10 50605065 1 5 89 5 999999999 60506056 1000000000

output

Copy

9999 9 50605065 1 5 89 5 999999999 60506055 999999999

Note

The first example is described in the problem statement.

题解:水题。。其实只要把偶数减一,奇数不变即可;

AC代码为:

#include<bits/stdc++.h>

using namespace std;

int main()

{

    ios::sync_with_stdio(false);

    cin.tie(0);

    int n,num[1010];

    cin>>n;

    for(int i=0;i<n;i++)

    {

        cin>>num[i];

        if((num[i]&1) ==0) num[i]-=1;

    }

    for(int i=0;i<n;i++) cout<<num[i]<<" ";

    cout<<endl;

    return 0;

}

CodeForces1006A - Adjacent Replacements的更多相关文章

  1. CodeForces Round #498 Div.3 A. Adjacent Replacements

    http://codeforces.com/contest/1006/problem/A Mishka got an integer array aa of length nn as a birthd ...

  2. Codeforces Div3 #498 A-F

                                                                               . A. Adjacent Replacement ...

  3. Codeforces Round #498 (Div. 3) 简要题解

    [比赛链接] https://codeforces.com/contest/1006 [题解] Problem A. Adjacent Replacements        [算法] 将序列中的所有 ...

  4. 【CSS3】Advanced3:Universal, Child, and Adjacent Selectors

    1.Universal selectors eg:#target*{ } 2.Child selectors < something immediately nested within some ...

  5. [CC-ADJLEAF2]Adjacent Leaves

    [CC-ADJLEAF2]Adjacent Leaves 题目大意: 给定一棵有根树,考虑从根开始进行DFS,将所有叶子按照被遍历到的顺序排列得到一个序列. 定义一个叶子集合合法,当且仅当存在一种DF ...

  6. Adjacent Bit Counts(01组合数)

    Adjacent Bit Counts 4557 Adjacent Bit CountsFor a string of n bits x 1 , x 2 , x 3 ,..., x n , the a ...

  7. 微软BI 之SSRS 系列 - 使用 LookupSet 和 Adjacent Group 等高级技巧在报表中跨 Dataset 分组查询

    SSRS 报表中有一些高级的技巧,平常很少用到,下面我通过这个案例来展现一下如何在实际开发中使用它们,并且如何解决一些实际的需求. 这张报表分别统计了不同的 Product 产品在不同的月份的 Ord ...

  8. nyoj 715 Adjacent Bit Counts

    描述 For a string of n bits x1, x2, x3, …, xn,  the adjacent bit count of the string  is given by      ...

  9. Adjacent Bit Counts(uvalive)

    For a string of n bits x1, x2, x3,…, xn, the adjacent bit count of the string (AdjBC(x)) is given by ...

随机推荐

  1. java map遍历方式及效率

    本文转载自Java Map遍历方式的选择. 只给出遍历方式及结论.测试数据可以去原文看. 如果你使用HashMap 同时遍历key和value时,keySet与entrySet方法的性能差异取决于ke ...

  2. 直接引用MrAdvice.dll文件不能实现AOP拦截,教你1分钟解决这个问题

    直接引用MrAdvice.dll文件不能实现AOP拦截,教你1分钟解决这个问题.近日工作中,要实现一个功能,那就是业务层方法里面实现自动缓存.编写业务的C#开发人员只关注如何将业务代码编写正确就可以了 ...

  3. js控制文本显示的字数,超出显示省略号

    在css中我们说了用css控制文本显示几行,超出用省略号,但这个办法是要完全占满一整行的,偏偏就是有UI设计师就不这么干,是不是很想打他,哈哈哈,我不会的时候都这样子在心里骂设计师的,这么久那么爱搞特 ...

  4. oracle实现"limit"功能

    转载于http://blog.sina.com.cn/s/blog_67e2758d0100s3oc.html oracle数据库不支持mysql中limit功能,但可以通过rownum来限制返回的结 ...

  5. 编译带libev和libuv的libwebsocket (Win平台)

    编译使用libwebsockets-2.0-stable版本. 编译依赖项目cmake,lws项目配置和生成依赖python, gpy工具依赖gpy, libuv项目配置和生成依赖,又一个项目配置和生 ...

  6. usaco training <1.2 Your Ride Is Here>

    题面 Your Ride Is Here It is a well-known fact that behind every good comet is a UFO. These UFOs often ...

  7. HTML的标签认识

    <!--      html标签     h1~h6 标题标签(只有1~6,依次减小)     p 段落标签     span 无意义的行标签     div 无意义的块标签     b 加粗 ...

  8. MySQL通过自定义函数实现递归查询父级ID或者子级ID

    背 景: 在MySQL中如果是有限的层次,比如我们事先如果可以确定这个树的最大深度, 那么所有节点为根的树的深度均不会超过树的最大深度,则我们可以直接通过left join来实现. 但很多时候我们是无 ...

  9. SCAU-1076 K尾相等数

    代码借鉴SCAU-OJ(感谢!!) 题目:1076 K尾相等数 时间限制:500MS  内存限制:65536K提交次数:251 通过次数:80 题型: 编程题   语言: G++;GCC   Desc ...

  10. Linux配置SSH和Xshell连接服务器

    >>>>>Ubuntu安装和配置ssh教程 SSH分为客户端 openssh-client 和服务器 openssh-server,可以利用以下命令确认电脑 上是否安装了 ...