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  • HDU 1517 A Multiplication Game (SG函数找规律)

    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;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/8427869.html
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