hdu 1517 A Multiplication Game

A Multiplication Game

 

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3113    Accepted Submission(s): 1785

Problem Description

Stan and Ollie play the game of multiplication by multiplying an integer p by one of the numbers 2 to 9. Stan always starts with p = 1, does his multiplication, then Ollie multiplies the number, then Stan and so on. Before a game starts, they draw an integer 1 < n < 4294967295 and the winner is who first reaches p >= n.

 

Input

Each line of input contains one integer number n.

 

Output

For each line of input output one line either 

Stan wins. 

or 

Ollie wins.

assuming that both of them play perfectly.

 

Sample Input

1621734012226

 

Sample Output

Stan wins.Ollie wins.Stan wins.
 
 博弈问题：
一：  2<=n<=9，第一个人赢；
二：  10<=n,第二个人赢；（因为可以取2到9这几个数，第一个人无论取哪个数都不会赢；所以，第一个人尽量取小的数字2，尽量使第二个  人不能赢，而第二个人最大取到9；因此，从10到2*9=18都是第一个人输。）
三：  18<n,（注意）第一个人只要取过之后使n/9落在10到18这个范围，那么第二个人就必输（相当于第一个人在10<=n<=18时一样），此时，n可以取到18*9；
四： 然后总结规律：第一个人输的区间为；[18^k*9+1,18^(k+1)](k>=0,k=0,1,2...)
 
#include <stdio.h> #include<math.h> #include<string.h>#define N 50int main() {     int t,flag;    __int64 n;    while(scanf("%I64d",&n)!=-1)    {        t=0;        flag=1;        while(n>=(__int64)pow(18,t)*9+1)        {            if(n<=(__int64)pow(18,t+1))                {                printf("Ollie wins.\n");                flag=0;                break;            }            t++;        }        if(flag==1)            printf("Stan wins.\n");    }    return 0; }