HDU 1517 A
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.
题目大意：
输入n，S和O依次将p = 1 乘以一个2~9的数字，先令p的值大于等于n的人获胜
解题思路：
2 ~ 9内，为S的必胜态；10 ~ 18内，为O的必胜态；超过18，S就可以通过控制第一轮的取值来获胜（比如乘以2或乘以9等）。
所以可以得出每乘18就会循环一轮，所以只需要让n不断除以18，判断最后是落在1 ~ 9还是10 ~ 18。
注意：由于存在除不尽的情况，需要将n设置为double
代码：
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
using namespace std;
int main() {
double n;
while( ~scanf("%lf", &n) ) {
while(n > 18) n /= 18;
if(n >= 1 && n <= 9) printf("Stan wins.\n");
else printf("Ollie wins.\n");
}
return 0;
}