Sengoku still remembers the mysterious "colourful meteoroids" she discovered with Lala-chan when they were little. In particular, one of the nights impressed her deeply, giving her the illusion that all her fancies would be realized.
On that night, Sengoku constructed a permutation p1, p2, ..., pn of integers from 1 to n inclusive, with each integer representing a colour, wishing for the colours to see in the coming meteor outburst. Two incredible outbursts then arrived, each with n meteorids, colours of which being integer sequences a1, a2, ..., an and b1, b2, ..., bn respectively. Meteoroids' colours were also between 1 and ninclusive, and the two sequences were not identical, that is, at least one i (1 ≤ i ≤ n) exists, such that ai ≠ bi holds.
Well, she almost had it all — each of the sequences a and b matched exactly n - 1 elements in Sengoku's permutation. In other words, there is exactly one i (1 ≤ i ≤ n) such that ai ≠ pi, and exactly one j (1 ≤ j ≤ n) such that bj ≠ pj.
For now, Sengoku is able to recover the actual colour sequences a and b through astronomical records, but her wishes have been long forgotten. You are to reconstruct any possible permutation Sengoku could have had on that night.
The first line of input contains a positive integer n (2 ≤ n ≤ 1 000) — the length of Sengoku's permutation, being the length of both meteor outbursts at the same time.
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ n) — the sequence of colours in the first meteor outburst.
The third line contains n space-separated integers b1, b2, ..., bn (1 ≤ bi ≤ n) — the sequence of colours in the second meteor outburst. At least one i (1 ≤ i ≤ n) exists, such that ai ≠ bi holds.
Output n space-separated integers p1, p2, ..., pn, denoting a possible permutation Sengoku could have had. If there are more than one possible answer, output any one of them.
Input guarantees that such permutation exists.
5
1 2 3 4 3
1 2 5 4 5
1 2 5 4 3
5
4 4 2 3 1
5 4 5 3 1
5 4 2 3 1
4
1 1 3 4
1 4 3 4
1 2 3 4
In the first sample, both 1, 2, 5, 4, 3 and 1, 2, 3, 4, 5 are acceptable outputs.
In the second sample, 5, 4, 2, 3, 1 is the only permutation to satisfy the constraints.
题解：
题目描述有一点恶心，先讲一讲题意。
说白了就是给你两个数列a和b，要你找一个数列c，使得c与a和b都最多只有一个不同的数，这就是为什么第二组样例只能有一组解的原因。
思路就是一个一个找a和b相同的数直接放到c中，然后分别试一试两种情况就可以了。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
#include<ctime>
#include<stack>
#include<vector>
using namespace std;
int n,a[],b[],c[],vis[];
int cnt1,cnt2,cnt3,cnt4;
int main()
{
int i,j;
scanf("%d",&n);
for(i=; i<=n; i++)
{
scanf("%d",&a[i]);
}
for(i=; i<=n; i++)
{
scanf("%d",&b[i]);
}
memset(c,-,sizeof(c));
for(i=; i<=n; i++)
{
if(a[i]==b[i])
{
if(!vis[a[i]])
{
c[i]=a[i];
vis[a[i]]=;
}
}
}
for(i=; i<=n; i++)
{
if(c[i]==-)
{
if(!cnt1)cnt1=i;
else
{
cnt2=i;
break;
}
}
}
for(i=; i<=n; i++)
{
if(!vis[i])
{
if(!cnt3)cnt3=i;
else
{
cnt4=i;
break;
}
}
}
if(!cnt2)c[cnt1]=cnt3;
else
{
int ans1=,ans2=;
if(a[cnt1]!=cnt3)ans1++;
if(b[cnt1]!=cnt3)ans1++;
if(a[cnt2]!=cnt4)ans2++;
if(b[cnt2]!=cnt4)ans2++;
if(ans1==&&ans2==)
{
c[cnt1]=cnt3;
c[cnt2]=cnt4;
}
else
{
c[cnt2]=cnt3;
c[cnt1]=cnt4;
}
} for(i=; i<=n; i++)
cout<<c[i]<<' ';
return ;
}