题意描述:Stan和Ollie玩游戏,两人轮流,首先Stan开始把1乘以一个2到9之间的整数,接着Ollie再将前一个数乘以一个2到9之间的整数。最先使这个数不小于n的人获胜。
解题思路:与其说是博弈,不如说是找到了规律,S胜:2-9(0+2)——9;O胜:10-18 (9+1)——9*2;S胜:19-162 (2*9+1)——9*2*9;O胜:163-324 (9*2*9+1)——9*2*9*2;S胜:324-2916 (9*2*9*2+1)——9*2*9*2*9。
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
162 17 34012226Sample Output
Stan wins. Ollie wins. Stan wins.
AC代码
#include<stdio.h>
#include<string.h>
int main()
{
int a[25],n,i;
while(~scanf("%d",&n))
{
int ans=0;
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
ans=ans^a[i];
}
if(ans==0)
printf("No\n");
else
printf("Yes\n");
}
return 0;
}