PAT Advanced 1069 The Black Hole of Numbers (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 (.

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

这道题主要考察了字符串的转换，我们使用转换字符串的方式，可以快速得到答案，

使用do while循环，避免了0或刚开始是6767不输出情况

#include <iostream>
#include <algorithm>
using namespace std;
int main() {
    int N;
    scanf("%d", &N);
    string str, rev;
    do {
        str = to_string(N);
        while(str.length() != 4) str = "0" + str;
        sort(str.begin(), str.end());
        rev = str;
        reverse(rev.begin(), rev.end());
        N = stoi(rev) - stoi(str);
        printf("%s - %s = %04d\n", rev.c_str(), str.c_str(), N);
    }while(N != 0 && N != 6174);
    return 0;
}