PAT-A-1069-he 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 (0,104).

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:

Sample Output 1:

7766 - 6677 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174

Sample Input 2:

Sample Output 2:

1	2222 - 2222 = 0000

Code

#include <iostream>
#include <algorithm>

using namespace std;

//sort默认是递增，因此只需要写一个递减
bool cmp_desc(int a,int b){
    return a>b;
}
//数组作为参数传递时：在函数中对数组元素的修改就等同于对原数组元素的修改
//数组可作为参数但不能作为返回值
//将数组转为数字法1
int to_num_1(int num[]){
    int res=0,i=4,product=1;
    for(i=3;i>=0;i--){
        res+=num[i]*product;
        product*=10;
    } 
    return res;
}
//将数组转为数字法2
int to_num_2(int num[]){
    int res=0;
    for(int i=0;i<4;i++)
        res=res*10+num[i];
    return res;
}

//将数字转为数组
void to_arr(int num,int res[]){
    for(int i=3;i>=0;i--){
        res[i]=num%10;
        num/=10;
    }
}

int main(){
	
	int n,MIN,MAX;
	scanf("%d",&n);
    int num[4];
    while(1){
        to_arr(n,num);
        sort(num,num+4);//递增排序
        MIN = to_num_1(num);
        sort(num,num+4,cmp_desc);//递减排序
        MAX = to_num_1(num);
        n=MAX-MIN;
        printf("%04d - %04d = %04d\n",MAX,MIN,n);
        //循环出口
        if(n==0||n==6174)break;
    }
	return 0;
}