1069 The Black Hole of Numbers （PAT甲级）

可以用do while来改进……

#include 
#include 
#include 

int main(){
    std::string a, b, tmp;
    std::cin >> a;
    while(a.size() < 4){
        a = "0" + a;
    }
    if(a == "6174"){
        sort(a.begin(), a.end());
        b = a;
        reverse(a.begin(), a.end());
        tmp = std::to_string(std::stoi(a) - std::stoi(b));
        std::cout << a << " - " << b << " = " << tmp << std::endl;
        return 0;
    }
    while(a != "6174" && a != "0000"){
        sort(a.begin(), a.end());
        b = a;
        reverse(a.begin(), a.end());
        tmp = std::to_string(std::stoi(a) - std::stoi(b));
        while(tmp.size() < 4){
            tmp = "0" + tmp;
        }
        std::cout << a << " - " << b << " = " << tmp << std::endl;
        a = tmp;
    }
    return 0;
}

题目如下：

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:

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