题解 | #小招喵跑步#

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

int move(int x) {
    // 处理负数
    if(x < 0) x = -x;
    // 处理小于等于3的情况
    if(x <= 3) return x;
    
    vector<int> dp(x + 1);
    // 初始化基础情况
    dp[0] = 0;
    dp[1] = 1;
    dp[2] = 2;
    dp[3] = 3;
    
    // 动态规划填表
    for(int i = 4; i <= x; i++) {
        if(i % 2 == 0) {
            dp[i] = dp[i/2] + 1;
        } else {
            dp[i] = min(dp[(i+1)/2] + 2, dp[i-1] + 1);
        }
    }
    
    return dp[x];
}

int main() {
    int x;
    cin >> x;
    cout << move(x);
    return 0;
}

import java.util.Scanner;

public class Main {
    public static int move(int x) {
        // 处理负数
        if(x < 0) x = -x;
        // 处理小于等于3的情况
        if(x <= 3) return x;
        
        int[] dp = new int[x + 1];
        // 初始化基础情况
        dp[0] = 0;
        dp[1] = 1;
        dp[2] = 2;
        dp[3] = 3;
        
        // 动态规划填表
        for(int i = 4; i <= x; i++) {
            if(i % 2 == 0) {
                dp[i] = dp[i/2] + 1;
            } else {
                dp[i] = Math.min(dp[(i+1)/2] + 2, dp[i-1] + 1);
            }
        }
        
        return dp[x];
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int x = sc.nextInt();
        System.out.println(move(x));
    }
}

def move(x):
    # 处理负数
    if x < 0:
        x = -x
    # 处理小于等于3的情况
    if x <= 3:
        return x
        
    dp = [0] * (x + 1)
    # 初始化基础情况
    dp[1] = 1
    dp[2] = 2
    dp[3] = 3
    
    # 动态规划填表
    for i in range(4, x + 1):
        if i % 2 == 0:
            dp[i] = dp[i//2] + 1
        else:
            dp[i] = min(dp[(i+1)//2] + 2, dp[i-1] + 1)
            
    return dp[x]

x = int(input())
print(move(x))