题意:两个玩家玩一个游戏,从 p = 1,开始,然后依次轮流选择一个2 - 9的数乘以 p,问你谁先凑够 p >= n。
析:找规律,我先打了一下SG函数的表,然后就找到规律了
我找到的是:
1 - 9 Stan wins. 1 ~ 9
10 - 18 Ollie wins. 9+1 ~ 9*2
19 - 162 Stan wins. 9*2+1 ~ 9*2*9
163 - 324 Ollie wins. 9*2*9+1 ~ 9*2*9*2
325 - 2916 Stan wins. 9*2*9*2+1 ~ 9*2*9*2*9
2917 - 5832 Ollie wins. 9*2*9*2*9+1 ~ 9*2*9*2*9*2
规律就很明显了。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #include <assert.h> #include <bitset> #include <numeric> #define debug() puts("++++") #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) #define sz size() #define pu push_up #define pd push_down #define cl clear() #define lowbit(x) -x&x //#define all 1,n,1 #define FOR(i,x,n) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e17; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1000 + 10; const int maxm = 100 + 2; const LL mod = 100000000; const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1}; const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) { return r >= 0 && r < n && c >= 0 && c < m; } int main(){ unsigned int n; while(scanf("%d", &n) == 1){ LL p = 1; while(1){ p *= 9; if(p >= n){ puts("Stan wins."); break; } p *= 2; if(p >= n){ puts("Ollie wins."); break; } } } return 0; }