题目描述：

Every positive number can be presented by the exponential form.For example, 137 = 2^7 + 2^3 + 2^0。

Let's present a^b by the form a(b).Then 137 is presented by 2(7)+2(3)+2(0). Since 7 = 2^2 + 2 + 2^0 and 3 = 2 + 2^0 , 137 is finally presented by 2(2(2)+2 +2(0))+2(2+2(0))+2(0). 

Given a positive number n,your task is to present n with the exponential form which only contains the digits 0 and 2.

输入：

For each case, the input file contains a positive integer n (n<=20000).

输出：

For each case, you should output the exponential form of n an a single line.Note that,there should not be any additional white spaces in the line.

样例输入：

样例输出：

2(2(2+2(0))+2)+2(2(2+2(0)))+2(2(2)+2(0))+2+2(0)

代码：

#include <iostream>  
#include <math.h>  
using namespace std;  

void present(int n){  
    if(n == 0)  
        cout<<"0";  
    if(n == 1){  
        return;  
    }  
    if(n == 2){  
        cout<<"2";  
        return;  
    }  
    int i = 0;  
    int j = 0;  
    while(n){  
        for(i=0;i<18;i++){  
            if(pow(2,i) > n)  
                break;  
        }  
        j = i - 1;  
        if(j!=1)  
            cout<<"2(";  
        else  
            cout<<"2";  
        present(j);  
        if(j!=1)  
            cout<<")";  
        n -= pow(2,j);  
        if(n != 0)  
            cout<<"+";  
    }  
}  

int main(){  
    int n;  
    while(cin>>n){  
        present(n);  
        cout<<endl;  
    }  
    return 0;  
}

作者提醒：

本题从输出的结果就可以看出来，这是一道递归的例子。
对于当前的数，先确定它在2^i次方左右，这里的i先确定下来,即使得2^i<=n成立的最大的i,
然后继续求n-pow(2,i),直到n=0
对于指数，也是同样的代码，只不过这里要注意的是，递归的出口，当指数为0，为1，为2时，
作为最基本的情况，要单独列出。