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 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.
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.
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.
5
1 2 4 5 10
1 1 3 5 9
10
10000 10 50605065 1 5 89 5 999999999 60506056 1000000000
9999 9 50605065 1 5 89 5 999999999 60506055 999999999
代码:
#include <bits/stdc++.h>
using namespace std; int N;
int num[1010]; int main() {
scanf("%d", &N);
for(int i = 1; i <= N; i ++) {
scanf("%d", &num[i]);
if(num[i] % 2 == 0)
num[i] -= 1;
} for(int i = 1; i <= N; i ++) {
printf("%d", num[i]);
printf("%s", i != N ? " " : "\n");
}
return 0;
}
CodeForces Round #498 Div.3 A. Adjacent Replacements的更多相关文章
- 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 ...
- Codeforces Round #498 (Div. 3)--E. Military Problem
题意问,这个点的然后求子树的第i个节点. 这道题是个非常明显的DFS序: 我们只需要记录DFS的入DFS的时间,以及出DFS的时间,也就是DFS序, 然后判断第i个子树是否在这个节点的时间段之间. 最 ...
- Codeforces Round #498 (Div. 3)
被虐惨了,实验室里数十位大佬中的一位闲来无事切题(二,然后出了5t,当然我要是状态正常也能出5,主要是又热又有蚊子什么的... 题都挺水的.包括F题. A: 略 B: 找k个最大的数存一下下标然后找段 ...
- Codeforces Round #498 (Div. 3) E. Military Problem (DFS)
题意:建一颗以\(1\)为根结点的树,询问\(q\)次,每次询问一个结点,问该结点的第\(k\)个子结点,如果不存在则输出\(-1\). 题解:该题数据范围较大,需要采用dfs预处理的方法,我们从结点 ...
- Codeforces Round #498 (Div. 3) D. Two Strings Swaps (思维)
题意:给你两个长度相同的字符串\(a\)和\(b\),你可以将相同位置上的\(a\)和\(b\)的字符交换,也可以将\(a\)或\(b\)中某个位置和对应的回文位置上的字符交换,这些操作是不统计的,你 ...
- Codeforces Round #486 (Div. 3) E. Divisibility by 25
Codeforces Round #486 (Div. 3) E. Divisibility by 25 题目连接: http://codeforces.com/group/T0ITBvoeEx/co ...
- Codeforces Round #486 (Div. 3) D. Points and Powers of Two
Codeforces Round #486 (Div. 3) D. Points and Powers of Two 题目连接: http://codeforces.com/group/T0ITBvo ...
- Codeforces Round #297 (Div. 2)D. Arthur and Walls 暴力搜索
Codeforces Round #297 (Div. 2)D. Arthur and Walls Time Limit: 2 Sec Memory Limit: 512 MBSubmit: xxx ...
- Codeforces Round #422 (Div. 2)
Codeforces Round #422 (Div. 2) Table of Contents Codeforces Round #422 (Div. 2)Problem A. I'm bored ...
随机推荐
- warning: remote HEAD refers to nonexistent ref, unable to checkout.解决
git branch -r origin/branch origin/hexo git checkout -b hexo origin/hexo
- axiospost请求向后端提交数据
Axios向后端提交数据容易接收不到原因是传参方式是request payload,参数格式是json,而并非用的是form传参,所以在后台用接收form数据的方式接收参数就接收不到了.post表单请 ...
- Jmeter压力测试工具基本使用
转:https://blog.csdn.net/envyfan/article/details/42715779
- vue2.0父子组件以及非父子组件通信
官网API: https://cn.vuejs.org/v2/guide/components.html#Prop 一.父子组件通信 1.父组件传递数据给子组件,使用props属性来实现 传递普通字符 ...
- window10启用administrator 和启用组策略编辑器
1,启用administrator账户 net user administrator /active:yes 2,启用组策略编辑器 新建一个文本文件.把下面代码粘贴进去.修改后缀名为.cmd ...
- java算法面试题:设计一个快速排序。双路快速排序,简单易于理解。
package com.swift; import java.util.ArrayList; import java.util.Collections; import java.util.Compar ...
- for循环语句中的先后执行顺序
for(int i=0;i<10;i++){ cout<<i; } 分析程序运行结果:for(cout<<"a";cout<<" ...
- 虚拟机Linux_Mint中安装vmtools增强工具
一开始用VmwarePro安装Linux系统时,系统的整体界面会缩在屏幕中间的一小块区域内.如图: 看的会非常吃力.为了更好的解决这个问题,就需要安装Vmtools增强工具.安装步骤如下: 1. ...
- Python_列表、字典、字符串、集合操作
一.list Python内置的一种数据类型是列表:list.list是一种有序的集合,可以随时添加和删除其中的元素.对于list的操作,我们要学会增删改查. 查 我们可以直接索引查找,也可以通过切片 ...
- 获取页面URL参数值
JavaScript function GetParams(urlAddress) { var i, strLength, str, keyName, keyValue, params = {}, u ...