1653: Champion of the Swordsmanship

1653: Champion of the Swordsmanship

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 11  Solved: 8
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Description

In Zhejiang University, there is a famous BBS named Freecity. Usually we call it 88.

Recently some students at the Humour board on 88 create a new game - Swordsmanship. Different from the common sword fights, this game can be held with three players playing together in a match. Only one player advances from the match while the other two are eliminated. Sometimes they also hold a two-player match if needed, but they always try to hold the tournament with as less matches as possible.

（最近，一些学生在幽默板上创造了一个新的游戏——剑术。与普通的剑战不同，这款游戏可以与三名玩家在一场比赛中一起玩。只有一名球员在比赛中取得进步，而另外两名则被淘汰。有时，如果需要的话，他们也会举办双人赛，但他们总是尽量少赛。）

Input

The input contains several test cases. Each case is specified by one positive integer n (0 < n < 1000000000), indicating the number of players. Input is terminated by n=0.

Output

For each test case, output a single line with the least number of matches needed to decide the champion.

Sample Input

3
4
0

Sample Output

1
2

#include<stdio.h>
int main()
{
    int n,m,k;
    while(scanf("%d",&n)&&n!=0)
    {
        if(n%2==0)
            printf("%d\n",n/2);
        else
            printf("%d\n",(n-1)/2);
    }
}