题目：链接
题目：
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
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.
Stan wins. or Ollie wins.
assuming that both of them play perfectly.
Sample Input
162 17 34012226
Sample Output
Stan wins. Ollie wins. Stan wins.题目意思是说：两个人乘数，给出的数从1开始，两个人乘的时候可以给原数乘以2到9之间的任何数，现在给出一个数n，判断在两者都采取最优策略的情况下，谁先能够使得乘完之后的数大于等于，谁就胜利。。。。解题思路： 如果输入是 2 ~ 9 ，因为Stan 是先手，所以Stan 必胜 如果输入是 10~18 ，因为Ollie 是后手，不管第一次Stan 乘的是什么，Stan肯定在 2 ~ 9 之间， 如果Stan乘以 2 ，那么Ollie就乘以 9 ，就到18了，如果Stan乘以 9 ， 那么Ollie乘以大于1的数都都能超过 10 ~ 18 中的任何一个数。Ollie 必胜 如果输入是 19 ~ 162，那么这个范围是 Stan 的必胜态 如果输入是 163 ~ 324 ，这是又是Ollie的必胜态 ............ 必胜态是对称的！！！ 双方都很聪明，所以这样胜负就决定于Ｎ了，如果Ｎ不断除 １８后的得到不足１８的数Ｍ，如果１＜Ｍ＜＝９则先手胜利，即Stan wins．如果９＜Ｍ＜＝１８ 则后手胜利．这样就直接对N进行除18，得到进行判断即可...上代码：#include <iostream> #include <cstdio> #include <string> #include <string.h> #include <map> #include <vector> #include <cstdlib> #include <cmath> #include <algorithm> #include <cmath> #include <queue> #include <set> #include <stack> using namespace std; int min(int a, int b) { if(a<=b) return a; return b; } int max(int a, int b) { if(a>=b) return a; return b; } double min(double a, double b) { if(a<=b) return a; return b; } double max(double a, double b) { if(a>=b) return a; return b; } int main() { double n; while(scanf("%lf", &n) != EOF) { while(n>18) n /= 18; if(n <= 9) printf("Stan wins.\n"); else printf("Ollie wins.\n"); } return 0; }
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