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.
InputThe 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.
OutputOutput 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 NoteIn 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.
给两个数组（元素为1-n）a, b求一个数组c 使a,与b中的元素与c 恰好只有一个元素不同，思维的精简度，直接求出a中重复的那个元素，用没出现的元素代替
判断是否满足条件即可，不可能出现没有重复的情况因为这样会导致要改的元素不止一个；。。。写题目前一定要把问题想清楚，不要急着下手
#include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<cmath> #include <bits/stdc++.h> using namespace std; const int N = 3e5+10; typedef long long LL; int a[N], b[N],p1[N], p2[N], v1[N], v2[N]; int l1=0,l2=0; int main() { int n; scanf("%d", &n); memset(v1,0,sizeof(v1)); memset(v2,0,sizeof(v2)); for(int i=1; i<=n; i++) { scanf("%d", &a[i]); v1[a[i]]++; } int k1=0, k2=0; for(int i=1; i<=n; i++) { if(v1[i]==0) l1=i; if(v1[i]>1) l2=i; } for(int i=1;i<=n;i++) { if(v1[a[i]]>1) p1[k1++]=i; } for(int i=1; i<=n; i++) scanf("%d", &b[i]); int h1=p1[0], h2=p1[1], cnt=0; a[h1]=l1; for(int i=1;i<=n;i++) { if(a[i]!=b[i]) cnt++; } if(cnt==1) { for(int i=1; i<=n; i++) printf("%d%c",a[i],i==n?'\n':' '); return 0; } a[h1]=a[h2]; a[h2]=l1; for(int i=1; i<=n; i++) printf("%d%c",a[i],i==n?'\n':' '); return 0; }
