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 <iostream>
#include"stdio.h"
#include"stdlib.h"
#include"string.h"
#include"algorithm"
using namespace std;
bool compareChar(char c1, char c2){
if(c1<c2)
return false;
return true;
}
char * stringMinus(char *s1, char *s2){
for(int i=3;i>=0;i--){
if(s1[i]>=s2[i]){
s1[i] = '0'+(s1[i]-s2[i]);
}
else{
s1[i] = '0'+(s1[i]-s2[i]+10);
if(i>0)
s1[i-1] -= 1;
}
}
return s1;
} int main()
{
int n;
char s[5]="0000";
char incr[5];
scanf("%d",&n);
int i=0;
while(n){
int x = n%10;
s[3-i] = '0'+x;
n/=10;
i++;
}
while(strcmp(s,"0000")!=0){
sort(s,s+4);
strcpy(incr,s); sort(s,s+4,compareChar); printf("%s - %s = ",s,incr);
printf("%s\n",stringMinus(s,incr));
if(strcmp(s,"6174")==0){
break;
}
}
return 0;
}
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