PAT-A 1069. The Black Hole of Numbers

1069. The Black Hole of Numbers

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 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000

程序代码:

#include<stdio.h>
char* zeng(char z[]);
char* jian(char z[]);
void int_char(char a[],int n);
int char_int(char a[]);
int main()
{
int N;
char z[5];
char j[5];
int a=0,b=0;
char num[5];
scanf("%d",&N);
if(N==6174)
N=7641;
while(N!=6174)
{
int_char(num,N);
jian(num);
a=char_int(num);
zeng(num);
b = char_int(num);
N= a-b;
printf("%s ",jian(num));
printf("- %s = ",zeng(num));
if(N!=0)
printf("%04d",N);
else
{
printf("0000");
putchar('\n');
break;
}
putchar('\n');
}

return 0;
}
char* zeng(char z[])
{
int i=3,j=0;
char tmp;
for(i=3;i>0;i--)
for(j=0;j<i;j++)
{
if(z[j]>z[j+1])
{
tmp = z[j];
z[j]=z[j+1];
z[j+1]=tmp;
}
}
z[4]='\0';
return z;
}

char* jian(char z[])
{
int i=3,j=0;
char tmp;
for(i=3;i>0;i--)
for(j=0;j<i;j++)
{
if(z[j]<z[j+1])
{
tmp = z[j];
z[j]=z[j+1];
z[j+1]=tmp;
}
}
z[4]='\0';
return z;
}
void int_char(char a[],int n)
{
int i=0;
a[0]=n%10+'0';
a[1]=(n/10)%10+'0';
a[2]=(n/100)%10+'0';
a[3]=(n/1000)%10+'0';
a[4]='\0';
}
int char_int(char a[])
{
int sum=0;
sum =((a[0]-'0')*1000+(a[1]-'0')*100+(a[2]-'0')*10+(a[3]-'0'));
return sum;
}