E30571: 十进制整数的反码

bit manipulation, http://cs101.openjudge.cn/practice/E30571/

#include <array>
#include <iostream>

auto main() -> int {
  int n;
  std::array<int, 31> T = {0, 1, 1};
  std::cin >> n;
  if (n >= 1 && n <= 2) {
    std::cout << T[n] << "\n";
    return 0;
  }
  for (int i = 3; i <= n; i++) {
    T[i] = T[i - 1] + T[i - 2] + T[i - 3];
  }
  std::cout << T[n] << "\n";
  return 0;
}

共用时10min

简单的位运算

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;

int get(int n) {
	if (n == 0) return 1;
	int ans = 0;
	while (n) {
		n >>= 1;
		ans++;
	}
	return (1 << ans) - 1;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    cout << get(n) - n << endl;
	return 0;
}