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

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000

由于要排列数字，所以利用了数组存储每一位数字，利用了两个函数直接返回A,B值，利用一个函数看最后是否为6174若是则输出后结束，若不是则需要重新给数组赋值

代码如下：

#include<iostream>
#include<string>
#include<algorithm>
#include<sstream>

using namespace std;
int  a[4];
int return_b_value(){
	int A=0;
	for(int i=0;i<4;++i){
		int index=a[i];
		A=A*10+index;
	}
	return A;
}
int return_a_value(){
	int A=0;
	for(int i=3;i>=0;--i){
		int index=a[i];
		A=A*10+index;
	}
	return A;
}

int final_value(int A,int B){
	if(A-B==6174){
		return 1;
	}else{
		return 0;
	}
}

void change_a(int value){
	
	for(int i=3;i>=0;--i){
		a[i]=value%10;
		value/=10;
	}
}

int cmp1(int c,int d){
	return c<d;
}
int main(){	
	int n;
	scanf("%d",&n);
	change_a(n);
	int flag=0;
	for(int i=1;i<4;++i){
		if(a[0]!=a[i]){
			flag=1;break;
		}
	}
	if(flag==0){
		printf("%04d - %04d = 0000",n,n);
		return 0;
	}
	while(1){
		sort(a,a+4,cmp1);
		int A,B;
		A=return_a_value();
		B=return_b_value();
		if(final_value(A,B)){
			printf("%04d - %04d = %04d",A,B,A-B);
			break;
		}else{
			printf("%04d - %04d = %04d\n",A,B,A-B);
			change_a(A-B);
		}
	}
	return 0;
}