zoj 2722 Head-to-Head Match（数学思维）

题目链接：

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2722

题目描述：

　　Our school is planning to hold a new exciting computer programming contest. During each round of the contest, the competitors will be paired, and compete head-to-head. The loser will be eliminated, and the winner will advance to next round. It proceeds until there is only one competitor left, who is the champion. In a certain round, if the number of the remaining competitors is not even, one of them will be chosed randomly to advance to next round automatically, and then the others will be paired and fight as usual. The contest committee want to know how many rounds is needed to produce to champion, then they could prepare enough problems for the contest.

Input

The input consists of several test cases. Each case consists of a single line containing a integer N - the number of the competitors in total. 1 <= N <= 2,147,483,647. An input with 0(zero) signals the end of the input, which should not be processed.

Output

For each test case, output the number of rounds needed in the contest, on a single line.

Sample Input

8
16
15
0

Sample Output

3
4
4

/*问题 输入一个1到2147483647的整数，计算并输出两两配对的次数
解题思路 其实就是2的几次方的问题*/
#include<cstdio>
#include<cmath>

int main()
{
    int n,i;
    while(scanf("%d",&n),n != 0){
        for(i=0;i<32;i++){
            if(pow(2,i) >= n){
                printf("%d
",i);
                break;
            }
        }
    }
    return 0;
}