題目描述
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.
輸入描述:

c++實現：

#include<iostream>
using namespace std;

int f[16
 
];

void print(int n) {
    bool first = true;
    int num = n;
    for (int i = 15; i >= 0; i--) {
        if (f[i] <= num) {
            if (!first) cout << "+";
            else first = false;
            if (i == 1) cout << "2";
            else if (i == 0) cout << "2(0)";
            else
 
 {
                cout << "2(";
                print(i);
                cout << ")";
            }
            num -= f[i];
        }
    }
}

int main() {
    f[0] = 1;
    for (int i = 1; i < 16; i++) f[i] = f[i-1]*2;
    int n;
    while (cin >> n) {
        print(n);
    }
    return 0;
}