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 1:6767Sample Output 1:
7766 - 6677 = 1089 9810 - 0189 = 9621 9621 - 1269 = 8352 8532 - 2358 = 6174Sample Input 2:
2222Sample 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; }