PAT1069:The Black Hole of Numbers

Posted 2020-10-15 0kk470

1069. The Black Hole of Numbers (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

思路

逻辑数学题
注意6174本身的判断和1111系列数字的判断，另外对于小于4位的数字也要进行插‘0‘处理。

代码

#include<iostream>
#include<algorithm>
#include<sstream>
using namespace std;


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

bool isEquation(string s1)
{
   int a = stoi(s1);
   if( a == 0 || a % 1111 == 0)
    return true;

   return false;
}

int main()
{
    string s;
    while(cin >> s)
    {
       if(isEquation(s))
       {
           cout << s << " - " << s << " = 0000" << endl;
           break;
       }
       if(s == "6174")
       {
           cout << "7641 - 1467 = 6174" << endl;
           break;
       }
       int num1 = 0,num2 = 0;
       while( num1 != 6174)
       {
           while(s.size() < 4)
             s.push_back(‘0‘);
           sort(s.begin(),s.end(),cmp);
           num1 = stoi(s);
           sort(s.begin(),s.end());
           num2 = stoi(s);
           printf("%04d - %04d = %04d\n",num1,num2,num1 - num2);
           num1 -= num2;
           s = to_string(num1);
       }
    }
}