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  • 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;
    }
    
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  • 原文地址:https://www.cnblogs.com/zxyqzy/p/10453659.html
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