题解 | #2的幂次方#

#include <iostream>
#include <vector>
using namespace std;

void er(int x) {
    vector<int> nums;
    int k = 0;
    while (x > 0) {
        if (x % 2)
            nums.push_back(k);
        k++;
        x /= 2;
    }
    for (int i = nums.size()-1; i > 0; i--) {

        if (nums[i] == 1) {
            cout << 2 << '+';
            continue;
        }
        else if (nums[i] == 0) {
            cout << "2(0)" << '+';
            continue;
        }
        cout << 2 << '(';
        er(nums[i]);
        cout << ')';
        cout << '+';

    }
    if (nums[0] == 1) {
        cout << 2 ;
     
    }
     else if (nums[0] == 0) {
            cout << "2(0)" ;
       
        }
   else{cout << 2 << '(';
    er(nums[0]);
    cout << ')';
    } 
}

int main() {
    int n;
    cin >> n;
er(n);
}
// 64 位输出请用 printf("%lld")