pat-a1069. The Black Hole of Numbers (20)

水题，以前用的string，发现sort也能用于char数组

#include<cstdio>#include<algorithm>#include<cstring>#include<functional>using namespace std;int main(){char a[6];int m,n,t;scanf("%s",a);int len=strlen(a);for(int i=len;i<4;++i) a[i]='0';do{sort(a,a+4,greater<int>());sscanf(a,"%d",&m); printf("%s -",a);sort(a,a+4);sscanf(a,"%d",&n);printf(" %s = %04d\n",a,m-n);t=m-n;sprintf(a,"%04d",t);}while(t!=6174&&t!=0);return 0;}

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