1 second
256 megabytes
standard input
standard output
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 n inclusive, 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.
解体思路：
由题可知，p中不能有重复的，而且数据必合理，那么只有两种情况：
1. 只在一个下标上不同，这时候只要将a[i]此时的值替换为之前从未在a数组中出现的数字便可；
2.在两个下标上不同，只需交换一次ab即可满足，这时必有两种情况:
(1).a第一个下标代表的数是重复的，此时b必不重复,那么就和b交换,但这时有个特殊情况那就是，和第一个下标重复的正好是第二个下标，进行判断：
如果第一个下标下的代表的数字在a中没出现过那么就交换第一个下标的ab,否则交换第二个下标中的ab；
(2).若a第一个下标代表的数不是重复的,此时只需交换第二个下标即可；
实现代码:
#include<bits/stdc++.h>
using namespace std; int main()
{
int t,i,a[],b[],c[],flag[];
cin>>t;
for(i=;i<=t;i++)
flag[i] = ;
for(i=;i<t;i++){
cin>>a[i];
flag[a[i]]++;
}
for(i=;i<t;i++)
cin>>b[i];
int cnt = ;
for(i=;i<t;i++){
if(a[i]!=b[i]){
c[cnt] = i;
cnt++;
}
}
if(cnt==){
for(i=;i<=t;i++){
if(flag[i]==){
a[c[]] = i;
}
}
}
else{if(flag[a[c[]]]>&&a[c[]]!=a[c[]])
a[c[]] = b[c[]];
else if(flag[a[c[]]]>&&a[c[]]==a[c[]]){
if(flag[b[c[]]]==)
a[c[]] = b[c[]];
else
a[c[]] = b[c[]];
}
else
a[c[]] = b[c[]];
}
for(i=;i<t-;i++)
cout<<a[i]<<" ";
cout<<a[t-]<<endl;
return ;
}