• 1069 The Black Hole of Numbers


    For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in non-increasing order first, and then in non-decreasing order, a new number can be obtained by taking the second number from the first one. Repeat in this manner we will soon end up at the number 6174 -- the black hole of 4-digit numbers. This number is named Kaprekar Constant.

    For example, start from 6767, we'll get:

    7766 - 6677 = 1089
    9810 - 0189 = 9621
    9621 - 1269 = 8352
    8532 - 2358 = 6174
    7641 - 1467 = 6174
    ... ...
    
     

    Given any 4-digit number, you are supposed to illustrate the way it gets into the black hole.

    Input Specification:

    Each input file contains one test case which gives a positive integer N in the range (.

    Output Specification:

    If all the 4 digits of N are the same, print in one line the equation N - N = 0000. Else print each step of calculation in a line until 6174 comes out as the difference. All the numbers must be printed as 4-digit numbers.

    Sample Input 1:

    6767
    
     

    Sample Output 1:

    7766 - 6677 = 1089
    9810 - 0189 = 9621
    9621 - 1269 = 8352
    8532 - 2358 = 6174
    
     

    Sample Input 2:

    2222
    
     

    Sample Output 2:

    2222 - 2222 = 0000

    思路:

      注意特殊情况0000 和 6174.

    Code:

     1 #include <bits/stdc++.h>
     2 
     3 using namespace std;
     4 
     5 int charToInt(vector<char>& v) {
     6     int num = 0;
     7     for (int i = 0; i < 4; ++i) num = num * 10 + v[i];
     8     return num;
     9 }
    10 
    11 vector<char> intTochar(int n) {
    12     vector<char> num(4);
    13     for (int i = 3; i >= 0; --i) {
    14         num[i] = n % 10;
    15         n /= 10;
    16     }
    17     return num;
    18 }
    19 
    20 int main() {
    21     int n;
    22     cin >> n;
    23     if (n == 0) printf("%04d - %04d = %04d
    ", 0, 0, 0);
    24     if (n == 6174) printf("%04d - %04d = %04d
    ", 7641, 1467, 6174);
    25     int last = -1, temp = n;
    26     while (last != temp) {
    27         vector<char> v = intTochar(temp);
    28         sort(v.begin(), v.end(), greater<char>());
    29         int max = charToInt(v);
    30         sort(v.begin(), v.end(), less<char>());
    31         int min = charToInt(v);
    32         last = temp;
    33         temp = max - min;
    34         if (temp == last) break;
    35         printf("%04d - %04d = %04d
    ", max, min, temp);
    36     }
    37     return 0;
    38 }
    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/12823840.html
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