﻿PAT(A) 1069. The Black Hole of Numbers (20)-Ocrosoft

# PAT(A) 1069. The Black Hole of Numbers (20)

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
```

```#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
#define strend string::npos
#define ms(a) memset(a,0,sizeof(a))
#define  rep(a,v,b) for(int a=v;a<b;a++)
#define  repe(a,v,b) for(int a=v;a<=b;a++)
typedef long long LL;
const LL LINF = LLONG_MAX / 2;
const int INF = INT_MAX / 2;
const int MAXN = 300000+ 10;
const int MOD = 1000000009;
int gcd(int a, int b)
{
if (!b)return a;
return gcd(b, a%b);
}
/*(◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕)*/
/*(◕‿‿◕) 签订契约，成为马猴烧酒吧! (◕‿‿◕)*/
/*(◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕)*/
using namespace std;
void solve(string s)
{
string x,y;
x=y=s;
sort(x.begin(),x.end(),greater<char>());
sort(y.begin(),y.end());
stringstream xx(x),yy(y);
int a,b;xx>>a;yy>>b;
if(a==b)
{
cout<<x<<" - "<<y<<" = "<<"0000"<<endl;
return;
}
int c=a-b;
string z="";
for(int i=0;i<4;i++)
{
z.push_back(c%10+'0');
c/=10;
}
reverse(z.begin(),z.end());
cout<<x<<" - "<<y<<" = "<<z<<endl;
if(z=="6174")return;
else solve(z);
}
int main()
{
string s;cin>>s;
while(s.size()!=4)
s.insert(s.begin(),'0');
solve(s);
return 0;
} ```