1069. The Black Hole of Numbers (20)

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 = 10899810 - 0189 = 96219621 - 1269 = 83528532 - 2358 = 6174

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000

题意：

给定一个四位的数字，写出该数字的黑洞过程

分析：

要求四位数字，注意在将数拆成各位的时候，注意不是求所有的为，因为有的数开始是0，所以只要四位，不然第二个，第三个测试点过不了

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#include <set>#include <map>using namespace std;int Min(int *d){int sum = 0;for(int i = 0;i < 4;i++)sum = sum*10+d[i];return sum;}int Max(int *d){int sum = 0;for(int i = 3;i >= 0;i--)   sum = sum*10+d[i];return sum;}int main(){    int n;    cin >> n;    while(1)    {    int d[4];    int g = 0;    for(int i = 0;i < 4;i++)    {    d[g++] = n%10;    n/=10;    }       sort(d,d+4);       int maxn = Max(d);       int minn = Min(d);       int mul = maxn - minn;       printf("%04d - %04d = %04d\n",maxn,minn,mul);       if(mul == 0||mul == 6174)       break;       n = mul;    }return 0;}

下面给出大神的超级短的代码：

#include <iostream>#include <algorithm>using namespace std;bool cmp(char a, char b) {return a > b;}int main() {    string s;    cin >> s;    s.insert(0, 4 - s.length(), '0');    do {        string a = s, b = s;        sort(a.begin(), a.end(), cmp);        sort(b.begin(), b.end());        int result = stoi(a) - stoi(b);        s = to_string(result);        s.insert(0, 4 - s.length(), '0');        cout << a << " - " << b << " = " << s << endl;    } while (s != "6174" && s != "0000");    return 0;}