zoj 2722 Head-to-Head Match（两两比赛）

Head-to-Head Match

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

 题意：两两比赛，胜出一人，胜者进入下一轮比赛。如果该轮总数是奇数，则随机选出一人直接参加下一轮比赛，求比赛总轮数。

#include <iostream>
#include <cmath>
using namespace std;
int main(){
    int n;
    while(cin >> n){
        if(n == 0)
            break;
        for(double i = 0; i < 32; i++){
            //pow函数的参数需要定义为double类型 
            if(pow(2, i) >= n){
                cout << i << endl;
                break;
            }
        }
    }
    return 0;
}