cf1072B. Curiosity Has No Limits(枚举)

题意

题目链接

给出两个序列\(a, b\)，求出一个序列\(t\)，满足

\[a_i = t_i | t_{i + 1}\]

\[b_i = t_i \& t_{i + 1}\]

同时,\(0 \leqslant a_i, b_i, t_i \leqslant 3\)

sol

打比赛的时候想到了拆位，从此走上不归路。。。

显然，当最后一位确定了之后，其他的数也都跟着确定了，。。

所以暴力枚举每个位置上是哪个数就行。。

/*
*/
#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long 
#define ll long long 
#define rg register 
#define pt(x) printf("%d ", x);
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 1e6 + 10, inf = 1e9 + 10, mod = 1e9 + 7, b = 1;
const double eps = 1e-9;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, a[maxn], b[maxn], f[maxn];
void solve(int val) {
    f[n] = val;
    for(int i = n - 1; i >= 1; i--) {
        bool flag = 0;
        for(int j = 0; j <= 3; j++) {
            int nx = f[i + 1]; f[i] = j;
            if((a[i] == (f[i] | f[i + 1])) && (b[i] == (f[i] & f[i + 1]))) {flag = 1; break;}
        }
        if(!flag) return ;
    }
    puts("yes");
    for(int i = 1; i <= n; i++) printf("%d ", f[i]);
    exit(0);
}
main() {
    //fin(a);
    n = read();
    for(int i = 1; i <= n - 1; i++) a[i] = read();
    for(int i = 1; i <= n - 1; i++) b[i] = read();

    solve(0); solve(1); solve(2); solve(3);
    puts("no");
    return 0;
}
/*

*/