1653: Champion of the Swordsmanship
Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 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); } }