PAT(A) 1069. The Black Hole of Numbers (20)

原题目：

原题链接：https://www.patest.cn/contests/pat-a-practise/1069

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 = 1089 9810 - 0189 = 9621 9621 - 1269 = 8352 8532 - 2358 = 6174 Sample Input 2:
2222
Sample Output 2:
2222 - 2222 = 0000

解题报告

题目是找出所有的数字黑洞。 需要注意的是输出顺序以及本身就是特殊数字的情况。

代码

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41

#include "stdio.h"
#include <algorithm>
#include <vector>
using namespace std;
int big(int num){
    vector<int> arr;
    arr.push_back(num / 1000);
    arr.push_back(num / 100 % 10);
    arr.push_back(num / 10 % 10);
    arr.push_back(num % 10);
    sort(arr.begin(),arr.end());
    return arr[3] * 1000 + arr[2] * 100 + arr[1] * 10 + arr[0];
}

int small(int num){
    vector<int> arr;
    arr.push_back(num / 1000);
    arr.push_back(num / 100 % 10);
    arr.push_back(num / 10 % 10);
    arr.push_back(num % 10);
    sort(arr.begin(),arr.end());
    return arr[0] * 1000 + arr[1] * 100 + arr[2] * 10 + arr[3];
}

int main(){
    int num = 0;
    scanf("%d",&num);
    int b = big(num);
    int s = small(num);
    int newnum = b - s;
    if (num == newnum)
        printf("%04d - %04d = %04d\n",b,s,newnum);
    while(num != newnum){
        printf("%04d - %04d = %04d\n",b,s,newnum);
        num = newnum;
        b = big(num);
        s = small(num);
        newnum = b - s;
    }
    //system("pause");
}