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

1069. The Black Hole of Numbers (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 = 10899810 - 0189 = 96219621 - 1269 = 83528532 - 2358 = 6174

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000

#include<iostream>#include<cstdio>#include<algorithm>#include<stdio.h>using namespace std;int a[4], b[4];void Fenli(int n){  a[0] = b[0] = n / 1000;  a[1] = b[1] = n % 1000 / 100;  a[2] = b[2] = n % 100 / 10;  a[3] = b[3] = n % 10;}int sum(int s[]){  int ans = 0;  for (int i = 0; i<4; i++)    ans = ans * 10 + s[i];  return ans;}int cmp(int a, int b){  return a > b;}int main(){  int min, max,ans;  cin >> ans;  Fenli(ans);  sort(a, a + 4);  sort(b, b + 4, cmp);  min = sum(a);  max = sum(b);  if (min == max||max-min==6174)  {    printf("%04d - %04d = %04d\n", max, min, max - min);    system("pause");    return 0;  }  while (ans != 6174)  {    Fenli(ans);    sort(a, a + 4);    sort(b, b + 4, cmp);    min = sum(a);    max = sum(b);    ans = max - min;    printf("%04d - %04d = %04d\n", max, min,ans);  }  system("pause");  return 0;    }