PAT甲级1069(数字黑洞)

1069 The Black Hole of Numbers (20 point(s))

  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 (0,10
​4
​​ ).

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:

222 - 2222 = 0000

题目解析

  唯一注意的是输入的不是4位数时在前面补0就行了

代码

#define _CRT_SECURE_NO_WARNINGS
#include 
#include 
#include 
using namespace std;

bool cmp(char a, char b) {
	return a > b;
}

//注意不一定输入4位数，不足4位要在前面补0
int main()
{
	string num, numDup;
	cin >> num;
	if (num.size() < 4) num.insert(0, string(4 - num.size(), '0'));
	numDup = num;	
	do {
		sort(num.begin(), num.end(), cmp);
		sort(numDup.begin(), numDup.end());
		int n1 = stoi(num), n2 = stoi(numDup);
		printf("%04d - %04d = %04d\n", n1, n2, n1 - n2);
		num = to_string(n1 - n2);
		if (num.size() < 4) num.insert(0, string(4 - num.size(), '0'));
		numDup = num;
	} while (num != "6174" && num != "0000");
	return 0;
}