B. Curiosity Has No Limits
When Masha came to math classes today, she saw two integer sequences of length n−1 on the blackboard. Let’s denote the elements of the first sequence as ai (0≤ai≤3), and the elements of the second sequence as bi (0≤bi≤3).
Masha became interested if or not there is an integer sequence of length n, which elements we will denote as ti (0≤ti≤3), so that for every i (1≤i≤n−1) the following is true:
ai=ti|ti+1 (where | denotes the bitwise OR operation) and
bi=ti&ti+1 (where & denotes the bitwise AND operation).
The question appeared to be too difficult for Masha, so now she asked you to check whether such a sequence ti of length n exists. If it exists, find such a sequence. If there are multiple such sequences, find any of them.
Input
The first line contains a single integer n (2≤n≤105) — the length of the sequence ti.
The second line contains n−1 integers a1,a2,…,an−1 (0≤ai≤3) — the first sequence on the blackboard.
The third line contains n−1 integers b1,b2,…,bn−1 (0≤bi≤3) — the second sequence on the blackboard.
Output
In the first line print “YES” (without quotes), if there is a sequence ti that satisfies the conditions from the statements, and “NO” (without quotes), if there is no such sequence.
If there is such a sequence, on the second line print n integers t1,t2,…,tn (0≤ti≤3) — the sequence that satisfies the statements conditions.
If there are multiple answers, print any of them.
Examples
inputCopy
4
3 3 2
1 2 0
outputCopy
YES
1 3 2 0
inputCopy
3
1 3
3 2
outputCopy
NO
Note
In the first example it’s easy to see that the sequence from output satisfies the given conditions:
t1|t2=(012)|(112)=(112)=3=a1 and t1&t2=(012)&(112)=(012)=1=b1;
t2|t3=(112)|(102)=(112)=3=a2 and t2&t3=(112)&(102)=(102)=2=b2;
t3|t4=(102)|(002)=(102)=2=a3 and t3&t4=(102)&(002)=(002)=0=b3.
In the second example there is no such sequence.
搜索题，dfs就过了
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int n,a[100005],b[100005],ans[100005],flag=0;
void dfs(int pos,int pre)
{
if(pos>n-1&&!flag)
{
flag=1;
printf("YES\n");
for(int i=0;i<=n-1;i++)
printf("%d ",ans[i]);
cout<<endl;
}
else
for(int i=0;i<=3;i++)
{
if(((pre|i)==a[pos])&&((pre&i)==b[pos])||(pos==n-1))
{
//printf("%d",i);
ans[pos]=pre;
dfs(pos+1,i);
}
}
}
int main()
{
//printf("%d %d\n",3|2,3&2);
scanf("%d",&n);
for(int i=0;i<n-1;i++)
scanf("%d",&a[i]);
for(int i=0;i<n-1;i++)
scanf("%d",&b[i]);
for(int i=0;i<=3;i++)
{
dfs(0,i);
if(flag)
break;
}
if(flag==0)
printf("NO\n");
return 0;
}