Curiosity Has No Limits
When Masha came to math classes today, she saw two integer sequences of length n−1n−1 on the blackboard. Let’s denote the elements of the first sequence as aiai (0≤ai≤3), and the elements of the second sequence as bibi (0≤bi≤3).
Masha became interested if or not there is an integer sequence of length nn, which elements we will denote as titi (0≤ti≤3), so that for every ii (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 titi of length nn 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 nn (2≤n≤1052≤n≤105) — the length of the sequence titi.
The second line contains n−1n−1 integers a1,a2,…,an−1a1,a2,…,an−1 (0≤ai≤3) — the first sequence on the blackboard.
The third line contains n−1n−1 integers b1,b2,…,bn−1b1,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 titi 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 nn integers t1,t2,…,tnt1,t2,…,tn (0≤ti≤30≤ti≤3) — the sequence that satisfies the statements conditions.
If there are multiple answers, print any of them.
Examples
input
4
3 3 2
1 2 0
output
YES
1 3 2 0
input
3
1 3
3 2
output
NO
Note
In the first example it’s easy to see that the sequence from output satisfies the given conditions:
- t1|t2=(01)|(11)=(11)=3=a1 and t1&t2=(01)&(11)=(01)=1=b1;
- t2|t3=(11)|(10)=(11)=3=a2 and t2&t3=(11)&(10)=(10)=2=b2;
- t3|t4=(10)|(00)=(10)=2=a3 and t3&t4=(10)&(00)=(00)=0=b3.
In the second example there is no such sequence.
题意：
给你a[],b[]，让你求出t[]. a[i]= t[i] | t[i+1]， b[i] = t[i] & t[i+1].满足这两条规则，且t[i]的范围只能说0，1，2。如果有解，输出"YES", 并输出t[i]。
题目链接：
http://codeforces.com/contest/1072/problem/B
思路：
dfs，先他给第一位的值，然后开始一位一位开始搜索， 当当前位满足条件时，搜索下一位，搜索到最后输出即可。
代码：
``
#include<bits/stdc++.h>
using namespace std;
int a[200000], b[200000];
int t;
int p[200000];
void dfs(int step, int k)
{
if(step == t) {
cout << "YES" << endl;
for(int i = 1; i <= t; i++)
cout << p[i] << " ";
cout << endl;
exit(0);
}
for(int i = 0; i <= 3; i++) {
if(((k | i) == a[step]) && ((k & i) == b[step])) {
p[step+1] = i;
dfs(step+1, i);
}
}
}
int main()
{
cin >> t;
for(int i = 1; i < t; i++)
cin >> a[i];
for(int i = 1; i < t; i++)
cin >> b[i];
for(int i = 0; i <= 3; i++) {
p[1] = i;
dfs(1, i);
}
cout << "NO" << endl;
return 0;
}
``