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.
比赛时比较匆忙，没有看清题意，好在C题刚做过类似的题，在没有优化下险过，不然和爆零没什么区别了
优化起来有点麻烦，就算咯~
继续说B题：
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.
这句话是说a数组和p数组仅有一个不同的元素，b数组也是。看懂了就简单啦！
遍历一遍a数组，找出缺少的那个元素t，再将a数组中重复的两个元素其中一个换成t，看一下和b数组是否满足题意即可。
#include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #include<iostream> #include<queue> #include<map> #include<cmath> #include<set> #include<stack> #define ll long long #define max(x,y) ((x)>(y)?(x):(y)) #define min(x,y) ((x)>(y)?(y):(x)) #define cls(name,x) memset(name,x,sizeof(name)) using namespace std; const int inf=1<<28; const int maxn=1010; const int maxm=2010; const int mod=1e9+7; const double pi=acos(-1.0); int n; int num1[maxn]; int num2[maxn]; bool judge(int a[],int b[]) { int c=0; for(int i=1;i<=n;i++) if(a[i]!=b[i]) c++; if(c==1) return true; else return false; } int main() { //freopen("in.txt","r",stdin); while(~scanf("%d",&n)) { for(int i=1;i<=n;i++) scanf("%d",&num1[i]); for(int i=1;i<=n;i++) scanf("%d",&num2[i]); int vis[maxn]; cls(vis,0); for(int i=1;i<=n;i++) vis[num1[i]]++; int t; for(int i=1;i<=n;i++) if(!vis[i]) t=i; for(int i=1;i<=n;i++) if(vis[num1[i]]==2) { int old=num1[i]; num1[i]=t; if(judge(num1,num2)) break; else num1[i]=old; } for(int i=1;i<=n;i++) printf("%d%s",num1[i],i==n?" ":" "); } return 0; }