Problem Statement
Given are two sequences A and B, both of length N. A and B are each sorted in the ascending order. Check if it is possible to reorder the terms of B so that for each i (1≤i≤N) Ai≠Bi holds, and if it is possible, output any of the reorderings that achieve it.
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Input
Input is given from Standard Input in the following format:
N
A1 A2 ⋯ AN
B1 B2 ⋯ BN
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Output
If there exist no reorderings that satisfy the condition, print ‘No’.
If there exists a reordering that satisfies the condition, print ‘Yes’ on the first line. After that, print a reordering of B on the second line, separating terms with a whitespace.
If there are multiple reorderings that satisfy the condition, you can print any of them.
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Sample Input 1
6
1 1 1 2 2 3
1 1 1 2 2 3
Sample Output 1
Yes
2 2 3 1 1 1
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Sample Input 2
3
1 1 2
1 1 3
Sample Output 2
No
题意
给定两个已经从小到大排列的A,B数组,是否可以对B作任意调换使得对于任意的i,Ai不等于Bi。
思路
既然已经从小到大排序了,则首先将B数组倒置,然后从前向后遍历,若寻到相同的就寻找B数组中是否有可以与此位交换后使得两位都不同的数存在,若有则交换,无则说明结果不存在。
代码
#include<map>
#include<stack>
#include<queue>
#include<string>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1e6+5;
int n,ans,res,sit,a[maxn],b[maxn];
int main()
{
//ios::sync_with_stdio(false);
//cin.tie(0);cout.tie(0);
cin>>n;
for(int i=1;i<=n;i++)
cin>>a[i];
for(int i=n;i>=1;i--)
cin>>b[i];
for(int i=1;i<=n;i++)
{
if(a[i]==b[i])
{
bool flag=0;
for(int j=1;j<=n;j++)
{
if(a[i]!=b[j]&&a[j]!=b[i])
{
swap(b[i],b[j]);
flag=1;
break;
}
}
if(!flag)
{
cout<<"No"<<endl;
return 0;
}
}
}
cout<<"Yes"<<endl;
for(int i=1;i<=n;i++)
cout<<b[i]<<" ";
cout<<endl;
return 0;
}