Coder Pat之路 1069 The Black Hole of Numbers

1069 The Black Hole of Numbers (20)（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, 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

User: 吴锦诚
Date: 2018/6/13

#include <iostream>
#include <algorithm>
#include <string>
#include <string.h>
using namespace std;
int up[4],down[4];
bool cmp(int a, int b) {
	return a > b;
}
int toInt(int ans[]) {
	return ans[0] * 1000 + ans[1] * 100 + ans[2] * 10 + ans[3];
}
int getup(int a) {
	up[3] = a % 10;
	up[2] = a / 10 % 10;
	up[1] = a / 100 % 10;
	up[0] = a / 1000;
	sort(up, up + 4);
	for (int i = 0; i < 4; i++) {
		cout << up[i];
	}
	cout << " = ";
	return toInt(up);
}
int getdown(int a) {
	down[3] = a % 10;
	down[2] = a / 10 % 10;
	down[1] = a / 100 % 10;
	down[0] = a / 1000;
	sort(down, down + 4, cmp);
	for (int i = 0; i < 4; i++) {
		cout << down[i];
	}
	cout << " - ";
	return toInt(down);
}
int subscribe(int n) {
	int a = getdown(n),b =  getup(n);
	int temp = a - b;
	up[3] = temp % 10;
	up[2] = temp / 10 % 10;
	up[1] = temp / 100 % 10;
	up[0] = temp / 1000;
	for (int i = 0; i < 4; i++) {
		cout << up[i];
	}
	cout << endl;
	return temp;
}
bool check(int a) {
	up[3] = a % 10;
	up[2] = a / 10 % 10;
	up[1] = a / 100 % 10;
	up[0] = a / 1000;
	if (up[0] == up[1] && up[1] == up[2]&& up[2] == up[3]){
		cout << a << " - " << a << " = " << "0000" << endl;
		return true;
	}
	return false;
}
int main()
{
	int n;
	while (cin>>n) {
		bool index = true;
		if (check(n)) {
			index = false;
		}
		while (index) {
			n = subscribe(n);
			if (n == 6174)
				break;
		}
		
	}
	return 0;
}