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 = 10899810 - 0189 = 96219621 - 1269 = 83528532 - 2358 = 61747641 - 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

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;}