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

刷乙级的时候我还不会qsort...

#include<stdio.h>int da(int x){int i,j,a[4];a[0]=x/1000;a[1]=x/100%10;a[2]=x/10%10;a[3]=x%10;for(i=0;i<3;i++){for(j=i;j<3;j++){if(a[i]<a[j+1]){int temp=a[i];a[i]=a[j+1];a[j+1]=temp;}}}return a[0]*1000+a[1]*100+a[2]*10+a[3];}int xiao(int x){int i,j,a[4];a[0]=x/1000;a[1]=x/100%10;a[2]=x/10%10;a[3]=x%10;for(i=0;i<3;i++){for(j=i;j<3;j++){if(a[i]>a[j+1]){int temp=a[i];a[i]=a[j+1];a[j+1]=temp;}}}return a[0]*1000+a[1]*100+a[2]*10+a[3];}int main(){int x;scanf("%d",&x);if(x/1000==x/100%10&&x/100%10==x/10%10&&x/10%10==x%10){printf("%d - %d = 0000",x,x);}else{int dd=da(x),xx=xiao(x);int cha;do{cha=dd-xx;printf("%04d - %04d = %04d\n",dd,xx,cha);dd=da(cha);xx=xiao(cha);}while(cha!=6174);}}