1069 The Black Hole of Numbers (20 分)（数学问题）

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

分析：

有一个测试用例注意点，如果当输入N值为6174的时候，依旧要进行下面的步骤，直到差值为6174才可以～所以用do while语句，无论是什么值总是要执行一遍while语句，直到遇到差值是0000或6174～

s.insert(0, 4 – s.length(), ‘0’);用来给不足4位的时候前面补0～

原文链接：https://blog.csdn.net/liuchuo/article/details/52497341

题解

#include <bits/stdc++.h>

using namespace std;

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n;
    cin>>n;
    string s=to_string(n);
    if(s[0]==s[1]&&s[1]==s[2]&&s[2]==s[3]){
        printf("%04d - %04d = 0000\n",n,n);
        return 0;
    }
    do
    {
        int a=0,b=0,i=0;
        vector<int> v(4);
        while(n)
        {
            v[i++]=n%10;
            n/=10;
        }
        sort(v.begin(),v.end());
        int sum=1;
        for(int i=0; i<v.size(); i++)
        {
            a+=v[i]*sum;
            sum*=10;
        }
        sum=1;
        for(int i=v.size()-1; i>=0; i--)
        {
            b+=v[i]*sum;
            sum*=10;
        }
        printf("%04d - %04d = %04d\n",a,b,a-b);
        n=a-b;
    }while(n!=6174);
    return 0;
}

柳神短小精悍的代码

膜QAQ

#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;
}