Description
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.
题意：连蒙带猜，大概说的是用两个序列还原出正确的序列，1-n各出现一次
解法：（读代码可能好懂许多）首先相同数字出现的一定要保存，然后不同的数字判断这两个数字以后会不会一定出现（即当中一个数字以后会在相同位置出现），出现过的以后不能再出现
于是。。。得到初始序列，不过这个序列还需要满足1-n各出现一次，所以还要保存未出现的数字，接下来就是按照题意规定在判断输出一次
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 const int maxn=123456; 5 int a,b,c; 6 int x[maxn],y[maxn],z[maxn]; 7 vector<int>q,p; 8 map<int,int>m,n,s,k; 9 int main() 10 { 11 cin>>a; 12 for(int i=1;i<=a;i++) 13 { 14 cin>>x[i]; 15 } 16 for(int i=1;i<=a;i++) 17 { 18 cin>>y[i]; 19 } 20 for(int i=1;i<=a;i++) 21 { 22 if(x[i]==y[i]) 23 { 24 s[x[i]]=1; 25 } 26 27 } 28 for(int i=1;i<=a;i++) 29 { 30 // cout<<s[x[i]]<<" "; 31 if(x[i]!=y[i]) 32 { 33 if(s[x[i]]==1)//相同位置不同数字，但x[i]在后面必须出现,选y[i] 34 { 35 k[y[i]]++; 36 q.push_back(y[i]); 37 } 38 else if(s[y[i]]==1)//相同位置不同数字，但y[i]在后面必须出现,选x[i] 39 { 40 k[x[i]]++; 41 q.push_back(x[i]); 42 } 43 else 44 { 45 if(k[x[i]]==0) 46 { 47 k[x[i]]++; 48 q.push_back(x[i]); 49 } 50 else if(k[y[i]]==0) 51 { 52 k[y[i]]++; 53 q.push_back(y[i]); 54 } 55 } 56 } 57 else 58 { 59 k[x[i]]++; 60 q.push_back(x[i]); 61 } 62 } 63 for(int i=0;i<q.size();i++) 64 { 65 z[q[i]]++; 66 } 67 for(int i=1;i<=a;i++) 68 { 69 if(z[i]==0) 70 { 71 p.push_back(i); 72 } 73 } 74 int ans=0; 75 map<int,int>o; 76 for(int i=0;i<q.size();i++) 77 { 78 if(x[i+1]==y[i+1]) 79 { 80 // z[x[i+1]]--; 81 cout<<x[i+1]<<" "; 82 } 83 else if(z[q[i]]==1) 84 { 85 cout<<q[i]<<" "; 86 } 87 else 88 { 89 z[q[i]]--; 90 cout<<p[ans++]<<" "; 91 } 92 } 93 return 0; 94 }