51 Nod 1067 博弈 SG函数
1067 Bash游戏 V2
有一堆石子共有N个。A B两个人轮流拿,A先拿。每次只能拿1,3,4颗,拿到最后1颗石子的人获胜。假设A B都非常聪明,拿石子的过程中不会出现失误。给出N,问最后谁能赢得比赛。
例如N = 2。A只能拿1颗,所以B可以拿到最后1颗石子。
输入
第1行:一个数T,表示后面用作输入测试的数的数量。(1 <= T <= 10000) 第2 - T + 1行:每行1个数N。(1 <= N <= 10^9)
输出
共T行,如果A获胜输出A,如果B获胜输出B。
输入样例
3 2 3 4
输出样例
B A A
1e9过大,打表找规律;
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<bitset> #include<ctime> #include<time.h> #include<deque> #include<stack> #include<functional> #include<sstream> //#include<cctype> //#pragma GCC optimize(2) using namespace std; #define maxn 200005 #define inf 0x7fffffff //#define INF 1e18 #define rdint(x) scanf("%d",&x) #define rdllt(x) scanf("%lld",&x) #define rdult(x) scanf("%lu",&x) #define rdlf(x) scanf("%lf",&x) #define rdstr(x) scanf("%s",x) #define mclr(x,a) memset((x),a,sizeof(x)) typedef long long ll; typedef unsigned long long ull; typedef unsigned int U; #define ms(x) memset((x),0,sizeof(x)) const long long int mod = 1e9 + 7; #define Mod 1000000000 #define sq(x) (x)*(x) #define eps 1e-5 typedef pair<int, int> pii; #define pi acos(-1.0) //const int N = 1005; #define REP(i,n) for(int i=0;i<(n);i++) typedef pair<int, int> pii; inline int rd() { int x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x; } ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); } int sqr(int x) { return x * x; } /*ll ans; ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans; } */ int f[maxn]; int SG[maxn], S[maxn]; void sg(int n) { // int i, j; ms(SG); for (int i = 1; i <= n; i++) { ms(S); for (int j = 1; f[j] <= i && j <= 3; j++) { S[SG[i - f[j]]] = 1; } for (int j = 0;; j++)if (!S[j]) { SG[i] = j; break; } } } int main() { // ios::sync_with_stdio(0); f[1] = 1; f[2] = 3; f[3] = 4; /* sg(100); int ans = 0; for (int i = 1; i <= 100; i++) { cout << i << ' '; ans = SG[i]; if (ans)cout << "A" << endl; else cout << "B" << endl; } */ int T = rd(); while (T--) { int n = rd(); if (n % 7 == 0 || ((n - 2) % 7 == 0)) { puts("B"); } else puts("A"); } return 0; }
EPFL - Fighting