(PAT 1069) The Black Hole of Numbers (简单数学)

- 日理万妓 2022-04-23 12:24 94阅读 0赞

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

解题思路:

将一个数的从大到小排序减去从小到大排序,值赋给这个数,依次循环直到得到6174或者0

这题的注意点是如果输入的数或者得到的值不足4位数的话,要用0补足

这题用string就可以解决,利用stoi和to\_string函数互相转换

#include <iostream>
    #include <algorithm>
    #include <string>
    using namespace std;
    
    bool cmpMAx(char a, char b) {
    	return a > b;
    }
    bool cmpMin(char a, char b) {
    	return a < b;
    }
    string num;
    int main() {
    	cin >> num;
    	int res = -1;
    	string maxsnum, minsnum;
    	while (true) {
    		if (num.length() < 4) {
    			for (int j = num.length(); j < 4; ++j) {
    				num += '0';
    			}
    		}
    		sort(num.begin(), num.end(), cmpMAx);
    		maxsnum = num;
    		sort(num.begin(), num.end(), cmpMin);
    		minsnum = num;
    		res = stoi(maxsnum) - stoi(minsnum);
    		printf("%04d - %04d = %04d\n", stoi(maxsnum), stoi(minsnum), res);
    		if(res == 0 || res == 6174) break;
    		num = to_string(res);
    	}
    	system("PAUSE");
    	return 0;
    }