1069 The Black Hole of Numbers
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 holeof 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
/*
Name:
Copyright:
Author: 流照君
Date: 2019/8/17 16:21:24
Description:
*/
#include <iostream>
#include<string>
#include <algorithm>
#include <vector>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
string s;
ll to_digit1(string ss)
{
ll n=ss.size();
ll sum=0;
for(int i=0;i<n;i++)
{
sum=sum*10+(ss[i]-'0');
}
return sum;
}
string to_string1 (ll sum1)
{
string ss;
while(sum1>0)
{
ll dig=sum1%10;
sum1=sum1/10;
ss=char(dig)+ss;
}
return ss;
}
bool cmp1(char x,char y)
{
return x<y;
}
bool cmp2(char x,char y)
{
return x>y;
}
int main(int argc, char** argv)
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
string s1,s2;
cin>>s;
s.insert(0, 4 - s.length(), '0');
s1=s;
s2=s;
sort(s1.begin() ,s1.end(),cmp1);
sort(s2.begin() ,s2.end(),cmp2);
if(s1==s2)
cout<<s2<<" - "<<s1<<" = "<<"0000"<<endl;
else
{
//ll ce=to_digit1("7856");
//cout<<ce<<endl;
//cout<<s2<<" - "<<s1<<" = "<<"0000"<<endl;
//exit(0);
do
{
int d1=stoi(s1);
int d2=stoi(s2);
int d3=d2-d1;
printf("%04d - %04d = %04d\n",d2,d1,d3);
//cout<<d2<<" - "<<d1<<" = "<<d3<<endl;
s=to_string(d3);
s.insert(0, 4 - s.length(), '0'); //the key
s1=s;
s2=s;
sort(s1.begin() ,s1.end(),cmp1);
sort(s2.begin() ,s2.end(),cmp2);
}while(s!="6174");
}
return 0;
}

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