PAT B1019/A1067 題解

1069 The Black Hole of Numbers （20 分）

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 (0,10

​4

​​ ).

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.

題目總體來說不難，隻要注意一點，在n為6174的時候，如果是用while循環中判斷的話，會導緻6174沒有輸出，是以必須在裡面進行判斷。還有一點就是格式錯誤…這個是真的看不出來其中是不是有空格，是以隻能自己測試…

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
bool cmp(int a,int b){
	return a>b;
}
int change_array(int a[]){
	int num=0;
	for(int i=0;i<4;i++){
		num=num*10+a[i];
	}
	return num;
}
int change_num(int a[],int n){
	int count=0;
	while(n!=0){
		a[count++]=n%10;
		n=n/10;
	}
}
int main(){
	int n,count=0,up=0,down=0;
	int num[5];
	cin>>n;
	change_num(num,n);
	sort(num,num+4);
	up=change_array(num);
	sort(num,num+4,cmp);
	down=change_array(num);
	while(true){
		n=down-up;
		printf("%04d - %04d = %04d\n",down,up,n);
		change_num(num,n);
		sort(num,num+4);
		up=change_array(num);
		sort(num,num+4,cmp);
		down=change_array(num);
		if(n==0||n==6174)
			break;
	}
	return 0;
}