cf1072B. Curiosity Has No Limits(列舉)
阿新 • • 發佈:2018-10-31
題意
給出兩個序列\(a, b\),求出一個序列\(t\),滿足
\[a_i = t_i | t_{i + 1}\]
\[b_i = t_i \& t_{i + 1}\]
同時,\(0 \leqslant a_i, b_i, t_i \leqslant 3\)
Sol
打比賽的時候想到了拆位,從此走上不歸路。。。
顯然,當最後一位確定了之後,其他的數也都跟著確定了,。。
所以暴力列舉每個位置上是哪個數就行。。
/* */ #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; } /* */