http://codeforces.com/contest/1072/problem/B
B. Curiosity Has No Limits
1 second
256 megabytes
standard input
standard output
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≤30≤ai≤3), and the elements of the second sequence as bibi (0≤bi≤30≤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≤30≤ti≤3), so that for every ii (1≤i≤n−11≤i≤n−1) the following is true:
- ai=ti|ti+1ai=ti|ti+1 (where || denotes the bitwise OR operation) and
- bi=ti&ti+1bi=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.
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≤30≤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≤30≤bi≤3) — the second sequence on the blackboard.
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.
4
3 3 2
1 2 0
YES
1 3 2 0
3
1 3
3 2
NO
In the first example it's easy to see that the sequence from output satisfies the given conditions:
- t1|t2=(012)|(112)=(112)=3=a1t1|t2=(012)|(112)=(112)=3=a1 and t1&t2=(012)&(112)=(012)=1=b1t1&t2=(012)&(112)=(012)=1=b1;
- t2|t3=(112)|(102)=(112)=3=a2t2|t3=(112)|(102)=(112)=3=a2 and t2&t3=(112)&(102)=(102)=2=b2t2&t3=(112)&(102)=(102)=2=b2;
- t3|t4=(102)|(002)=(102)=2=a3t3|t4=(102)|(002)=(102)=2=a3 and t3&t4=(102)&(002)=(002)=0=b3t3&t4=(102)&(002)=(002)=0=b3.
In the second example there is no such sequence.
题意：求一个n长度的t数组，满足t[i]&t[i+1] == b[i] 同时 t[i]|t[i+1] == a[i]。 a,b数组长度为n-1
枚举法，想过枚举但是没有想到是这样做的。因为t数组中，最后一个是最特殊的，只与a和b数组中的一个元素相关（自己没有发现）。同时没有发现如果确定了一个点，那么下一个点的值是唯一确定的。这样想来，其实枚举第一个也是可行的了。
同时，需要记住，如果t[i]确定了，那么与不同的t[i+1]进行或，与运算一定会得到的是不同的结果。
和之前做过的一道Fliptile有些像。
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <stack>
#define ll long long
//#define local using namespace std; const int MOD = 1e9+;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int maxn = (1e5+);
const int maxedge = *;
int a[maxn], b[maxn];
int t[maxn]; int main() {
#ifdef local
if(freopen("/Users/Andrew/Desktop/data.txt", "r", stdin) == NULL) printf("can't open this file!\n");
#endif
int n;
scanf("%d", &n);
for (int i = ; i < n-; ++i) {
scanf("%d", a+i);
}
for (int i = ;i < n-; ++i) {
scanf("%d", b+i);
}
for (int i = ; i < ; ++i) {
memset(t, -, sizeof(t));
t[n-] = i;
for (int j = n-; j >= ; --j) {
for (int k = ; k < ; ++k) {
if ((k|t[j+])==a[j] && (k&t[j+])==b[j]) {
t[j] = k;
break;
}
}
if (t[j] == -) break;
}
if (t[] != -) break;
}
if (t[] == -) printf("NO\n");
else {
printf("YES\n");
for (int i = ; i < n; ++i) {
printf("%d", t[i]);
if (i != n-) printf(" ");
}
printf("\n");
}
#ifdef local
fclose(stdin);
#endif
return ;
}