ALGO-14 VIP试题回文数（9.27）

柔情只为你懂 2022-12-11 02:30 167阅读 0赞

写两个函数就行。。。  
(好吧，重写了之后，函数又多了一个，但是感觉更和逻辑了)

#include <iostream>
    #include <cstdlib>
    #include <string>
    
    using namespace std;
    
    char table_hex[30] = {
        '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};
    int N, steps = 0;
    string M;//原字符串
    
    void string_reverse(string s1, string &targetstring)
    {
        
        for (int i = s1.size() - 1; i >= 0; i--)
        targetstring.push_back(s1[i]);
    }
    bool Is_Palindrome(string s)
    {
        
        string s_reverse;
        string_reverse(s, s_reverse);
        if (s == s_reverse) return true;
        else return false;
    }
    void machining(string &s)
    {
        
        string s_reverse;
        string_reverse(s, s_reverse);
        //carry表示进位
        int carry = 0;
    
        for (int i = s.size() - 1; i >= 0; i--)
        {
        
            int num1, num2;
            if (s[i] >= '0' && s[i] <= '9')
                num1 = s[i] - '0';
            else 
                num1 = N + (int)(s[i] - 'A');
            if (s_reverse[i] >= '0' && s_reverse[i] <= '9')
                num2 = s_reverse[i] - '0';
            else 
                num2 = N + (int)(s_reverse[i] - 'A');
            num1 += (num2 + carry);
            carry = num1 / N;
            num1 %= N;
            
            s[i] = table_hex[num1];
        }
        if (carry == 1) s = '1' + s;
        cout << '[' << s << ']' << endl;
    }
    int main()
    {
        
        ios::sync_with_stdio(false);
        cin >> N >> M;
        
        if (Is_Palindrome(M)) cout << "STEP=0" << endl;
        else 
        {
        
            while (true)
            {
        
                if (steps > 30) break;
                machining(M); steps++; //加工一次记数一次
                if (Is_Palindrome(M) ) break;
            }
            if (steps <= 30) cout << "STEP=" << steps << endl;
            else cout << "impossible" << endl;
        }
        system("pause");
        return 0;
    }