hdu 1517 A Multiplication Game

最新推荐文章于 2021-07-17 12:59:26 发布

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A Multiplication Game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6380 Accepted Submission(s): 3632

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

Sample Output

  
   Stan wins.
Ollie wins.
Stan wins.

题意：起始p=1，两个人轮流乘2~9,谁先到达终点就赢！

2~9 Stan必赢

10~18 Stan必输

19~27 Stan必赢

……

两人轮流有赢的段。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
int main() 
{
    int n,i;
    while(cin>>n)
    {
    	for(i=0;n>1;i++)
    	{
    		if(i&1)
    			n=ceil(n*1.0/2);
    		else
    			n=ceil(n*1.0/9);
    	}
    	if(i&1)
    		printf("Stan wins.\n");
    	else
    		printf("Ollie wins.\n");
    }
    return 0;
}

函数名： ceil

用法： double ceil(double x);

功能：返回大于或者等于指定表达式的最小整数

头文件： math.h

返回数据类型：double