【PAT】B1019/A1069. The Black Hole of Numbers (20)

Description:
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

//NKW 乙级真题1009
#pragma warning(disable:4996)
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
bool cmp(int a, int b){
	return a > b;
}
int toint(int a[]){
	int sum = 0;
	for (int i = 0; i < 4; i++)
		sum = sum * 10 + a[i];
	return sum;
}
void toarray(int a[], int b){
	for (int i = 3; i >= 0; i--){
		a[i] = b % 10;
		b /= 10;
	}
}
int main(){
	int n, max, min, arr[4];
	scanf("%d", &n);
	do{
		toarray(arr, n);
		sort(arr, arr + 4);
		min = toint(arr);
		sort(arr, arr + 4, cmp);
		max = toint(arr);
		n = max - min;
		printf("%04d - %04d = %04d\n", max, min, n);
	} while (n != 0 && n != 6174);
	system("pause");
	return 0;
}