题解 | #2的幂次方#
2的幂次方
https://www.nowcoder.com/practice/7cf7b0706d7e4b439481f53e5fdac6e7
#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")
