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  • A1069. 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 (0, 10000).

    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

     1 #include<cstdio>
     2 #include<iostream>
     3 #include<algorithm>
     4 using namespace std;
     5 bool cmp1(int a, int b){
     6     return a < b;
     7 }
     8 bool cmp2(int a, int b){
     9     return a > b;
    10 }
    11 void numSort(int n, int &r1, int &r2){
    12     int temp[20];
    13     int i = 0;
    14     r1 = 0; r2 = 0;
    15     do{
    16         temp[i++] = n % 10;
    17         n = n / 10;
    18     }while(n != 0 || i < 4);
    19     sort(temp, temp + i, cmp1);
    20     for(int j = 0, P = 1; j < i; j++){
    21         r1 = r1 + P * temp[j];
    22         P = P * 10;
    23     }
    24     sort(temp, temp + i, cmp2);
    25     for(int j = 0, P = 1; j < i; j++){
    26         r2 = r2 + P * temp[j];
    27         P = P * 10;
    28     }
    29 }
    30 int main(){
    31     int N, r1, r2, ans;
    32     scanf("%d", &N);
    33     numSort(N, r1, r2);
    34     do{
    35         ans = r1 - r2;
    36         printf("%04d - %04d = %04d
    ", r1, r2, ans);
    37         numSort(ans, r1, r2);
    38     }while(ans != 6174 && ans != 0);
    39     cin >> N;
    40     return 0;
    41 }
    View Code

    总结:

    1、注意在int转换为num[ ]数组时,如果不够四位,应补全成四位,否则答案会出错。(15应转换为0015和1500,而不是15和50)。

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  • 原文地址:https://www.cnblogs.com/zhuqiwei-blog/p/8514431.html
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