九度OJ 1095 2的幂次方

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：

1315

样例输出：

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

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。

#include 
#include 
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<