CodeForces An express train to reveries（简单的模拟）

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.

Input
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
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.

Examples
input
5
1 2 3 4 3
1 2 5 4 5
output
1 2 5 4 3
input
5
4 4 2 3 1
5 4 5 3 1
output
5 4 2 3 1
input
4
1 1 3 4
1 4 3 4
output
1 2 3 4
Note
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.
题目意思：给你两个数字序列，这两个数字序列至少有一位是不同的且数字范围都在1-n，且都有n-1个不同的数，要求输出一个新数列，包含n个数且和给定的两个数列只有一个数据不同。
解题思路：因为输出数列和给定的两个数列分别只有一个数据不同，所以两个数列最多只有两个数据不同，然后分两种情况输出给定数列即可。具体看代码

#include <iostream>
#include <bits/stdc++.h>

using namespace std;

int main()
{    int n,a[1005],b[1005],vis[1005],c[1005],d[3];///c数组来保存最后要输出的数据，d数组用来记录两个数列的不数据的下标
    while(~scanf("%d",&n)){
        memset(vis,0,sizeof(vis));
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<n;i++)
            scanf("%d",&b[i]);
        int k=0;///记录两个数列中有几个不同的数
        memset(c,0,sizeof(c));
        for(int i=0;i<n;i++)
        {
            if(a[i]==b[i])
            {
                 c[i] = a[i];
                 vis[c[i]]++;
            }
            else
                d[k++]=i;
        }
        if(k==2){///如果有两个数据不同
            vis[a[d[0]]]++;///分别将这两个数据标记
            vis[b[d[1]]]++;
            if(vis[a[d[0]]]==1&&vis[a[d[0]]]==vis[b[d[1]]])///如果不同的两个数据不相等，就让c数组的两个没赋值的值分别等于a[d[0]]，b[d[1]]或b[d[0]]，a[d[1]]
            {
                c[d[0]]=a[d[0]];
                c[d[1]]=b[d[1]];
            }
            else{
                c[d[0]]=b[d[0]];
                c[d[1]]=a[d[1]];
            }
        }
    else{///有一个数据不同
        for(int i=1;i<=n;i++){///找到那个缺少数据的并赋值给c[d[0]]
            if(vis[i]==0)
                c[d[0]]=i;
       }
    }
    for(int i=0;i<n-1;i++)///输出c数组
        printf("%d ",c[i]);
    printf("%d\n",c[n-1]);
    }
    return 0;
}