PAT刷题——黑洞数字

题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805400954585088

1069 The Black Hole of Numbers (20分)

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 (0,104).

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:

Sample Output 1:

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

Sample Input 2:

Sample Output 2:

2222 - 2222 = 0000

题目大意：https://pintia.cn/problem-sets/994805260223102976/problems/994805302786899968，与乙级题目 ”数字黑洞一样“一样

解题思路：将过程模拟出来就是了。首先需要输入一个n，表示开始的数字，建立一个while(n)循环,里面是n这样n为0的时候就可以跳出循环，结束循环的条件n为0或者n==6174，不然需要一直循环下去，直到满足条件。建立一个数组a[]来存储n中每一位的数字，一共四位，再将a[]从大到小排序，然后将a[]中的四个数字恢复成一个四位的整数，这样就得出第一个数字，同理，从小到大排序，再恢复成四位的整数，得到第二个数字，第二个数字见去第一个数字赋值给n，将这三个数字按照格数输出即可，n不满足条件继续循环。注意：输出的数字需要时四位，不足四位需要从前面填0，补成四位。

代码如下：

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
bool cmp(int a, int b) //从大到小排序
{
    return a > b;
}
int to_int(int s[]) //将每位数字合并成一个四位的整数
{
    int a = 0;
    for (int i = 0; i < 4; i++)
    {
        a = a * 10 + s[i];
    }
    return a;
}
int main()
{
    int n;
    int a[5]; //用来存储每位数字
    int max;  //存储从大到小排序之后的整数
    int min;  //从小到大排序之后的整数
    cin >> n;
    while (n) //n为0的时候直接退出循环
    {
        for (int i = 0; i < 4; i++) //将一个四位整数拆开，每一位分别存储到数组中
        {
            a[i] = n % 10;
            n /= 10;
        }
        sort(a, a + 4, cmp); //从大到小排序
        max = to_int(a);     //大数
        sort(a, a + 4);      //从小到大排序
        min = to_int(a);     //小数
        n = max - min;       //相减之后的数
        printf("%04d - %04d = %04d\n", max, min, n);
        if (n == 6174) //等于6174的时候退出
            break;
    }
    return 0;
}