CodeForce 517 Div 2. B Curiosity Has No Limits

http://codeforces.com/contest/1072/problem/B

B. Curiosity Has No Limits

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

When Masha came to math classes today, she saw two integer sequences of length $\mathrm{n-1n-1 on\; the\; blackboard.\; Let\text{'}s\; denote\; the\; elements\; of\; the\; first\; sequence\; as aiai (0\le ai\le 30\le ai\le 3),\; and\; the\; elements\; of\; the\; second\; sequence\; as bibi (0\le bi\le 30\le bi\le 3).}$

Masha became interested if or not there is an integer sequence of length $\mathrm{nn,\; which\; elements\; we\; will\; denote\; as titi (0\le ti\le 30\le ti\le 3),\; so\; that\; for\; every ii (1\le i\le n-11\le i\le n-1)\; the\; following\; is\; true:}$

• ${\mathrm{ai=ti|ti+1ai=ti|ti+1 (where || denotes\; the bitwise\; OR\; operation)\; and}}^{}$
• ${\mathrm{bi=ti\&ti+1bi=ti\&ti+1 (where \&\& denotes\; the bitwise\; AND\; operation).}}^{}$

The question appeared to be too difficult for Masha, so now she asked you to check whether such a sequence ${\mathrm{titi of\; length nn exists.\; If\; it\; exists,\; find\; such\; a\; sequence.\; If\; there\; are\; multiple\; such\; sequences,\; find\; any\; of\; them.}}^{}$

Input

The first line contains a single integer $\mathrm{nn (2\le n\le 1052\le n\le 105) —\; the\; length\; of\; the\; sequence titi.}$

The second line contains $\mathrm{n-1n-1 integers a1,a2,\dots ,an-1a1,a2,\dots ,an-1 (0\le ai\le 30\le ai\le 3) —\; the\; first\; sequence\; on\; the\; blackboard.}$

The third line contains $\mathrm{n-1n-1 integers b1,b2,\dots ,bn-1b1,b2,\dots ,bn-1 (0\le bi\le 30\le bi\le 3) —\; the\; second\; sequence\; on\; the\; blackboard.}$

Output

In the first line print "YES" (without quotes), if there is a sequence ${\mathrm{titi that\; satisfies\; the\; conditions\; from\; the\; statements,\; and\; "NO"\; (without\; quotes),\; if\; there\; is\; no\; such\; sequence.}}^{}$

If there is such a sequence, on the second line print $\mathrm{nn integers t1,t2,\dots ,tnt1,t2,\dots ,tn (0\le ti\le 30\le ti\le 3) —\; the\; sequence\; that\; satisfies\; the\; statements\; conditions.}$

If there are multiple answers, print any of them.

Examples

input

Copy

4
3 3 2
1 2 0

output

Copy

YES
1 3 2 0 

input

Copy

3
1 3
3 2

output

Copy

NO

Note

In the first example it's easy to see that the sequence from output satisfies the given conditions: 

• ${\mathrm{t1|t2=(012)|(112)=(112)=3=a1t1|t2=(012)|(112)=(112)=3=a1 and t1\&t2=(012)\&(112)=(012)=1=b1t1\&t2=(012)\&(112)=(012)=1=b1;}}^{}$
• ${\mathrm{t2|t3=(112)|(102)=(112)=3=a2t2|t3=(112)|(102)=(112)=3=a2 and t2\&t3=(112)\&(102)=(102)=2=b2t2\&t3=(112)\&(102)=(102)=2=b2;}}^{}$
• ${\mathrm{t3|t4=(102)|(002)=(102)=2=a3t3|t4=(102)|(002)=(102)=2=a3 and t3\&t4=(102)\&(002)=(002)=0=b3t3\&t4=(102)\&(002)=(002)=0=b3.}}^{}$

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+7;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int maxn = (1e5+10);
const int maxedge = 100*100;
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!
");
#endif
    int n;
    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 < 4; ++i) {
        memset(t, -1, sizeof(t));
        t[n-1] = i;
        for (int j = n-2; j >= 0; --j) {
            for (int k = 0; k < 4; ++k) {
                if ((k|t[j+1])==a[j] && (k&t[j+1])==b[j]) {
                    t[j] = k;
                    break;
                }
            }
            if (t[j] == -1) break;
        }
        if (t[0] != -1) break;
    }
    if (t[0] == -1) printf("NO
");
    else {
        printf("YES
");
        for (int i = 0; i < n; ++i) {
            printf("%d", t[i]);
            if (i != n-1) printf(" ");
        }
        printf("
");
    }
#ifdef local
    fclose(stdin);
#endif
    return 0;
}