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.
这题跟B19是一样的
#include <stdio.h> #include <stdlib.h> #include <algorithm> using namespace std; int Sort1(int a1,int a2,int a3,int a4); int Sort2(int a1,int a2,int a3,int a4); int main () { int a1=0,a2=0,a3=0,a4=0,i=0; int num; scanf("%d",&num); a1=num/1000; a2=(num/100)%10; a3=(num%100)/10; a4=num%10; if(a1==a2 && a2==a3 && a3==a4) { printf("%d%d%d%d - %d%d%d%d = 0000\n",a1,a1,a1,a1,a1,a1,a1,a1); return 0; } int result=0,first=0,second=0; while(1) { first=Sort2(a1,a2,a3,a4); second=Sort1(a1,a2,a3,a4); result=first-second; a1=result/1000; a2=(result/100)%10; a3=(result%100)/10; a4=result%10; printf("%04d - %04d = %04d\n",first,second,result); if(result==6174||result==0) break; } system("pause"); return 0; } int Sort1(int a1,int a2,int a3,int a4) { int a[4]={a1,a2,a3,a4}; sort(a,a+4); return (a[0]*1000+a[1]*100+a[2]*10+a[3]); } bool cmp(int a,int b) { return a>b; } int Sort2(int a1,int a2,int a3,int a4) { int a[4]={a1,a2,a3,a4}; sort(a,a+4,cmp); return (a[0]*1000+a[1]*100+a[2]*10+a[3]); }