PAT自主训练记录 甲级1069/乙级1019 The Black Hole of Numbers

A1069 The Black Hole of Numbers（20分）

题目描述

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.

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

题目大意

给一个不超过4位的数N，将其各个位上的数递减排，不妨记为MAX，再将其各个位上的数递增排，不妨记为MIN，用MAX-MIN得到一个新的数 $N_i$，若 $N_i$≠6174且 $N_i$≠0，则对 $N_i$进行同样的操作；若 $N_i$=0或6174则停止操作。同时，将每一步的减法运算输出。

思路
写一个函数将int型的数各个位分解存入数组，再写一个函数将数组形式的数转为int。写一个排序函数，对数组内的数进行排序，从而获得被减数和减数。while循环进行减法运算，判断是否满足输出停止的条件。

过程中的弯路或错误

• 将int数存入数组时，不需要按照原数位顺序存入，因为后面都是需要重新进行排序的。
• 输出格式用%04d可以保证都是以4位数的形式输出的。
• 一开始没注意输出格式要求数字与符号之间有空格。

最终正确代码

#include<stdio.h>
#include<cstdio>
#include<math.h>
#include<cmath>
#include <iostream>
#include<algorithm>
using namespace std;//递减排序
bool cmp(int a,int b){return a>b;
}//将整数n转化为数组里的数字 这里无所谓是否按照数位从高到低存入，只要把每一个数位上的数字存入数组即可
void to_array(int n,int a[]){for(int i=0;i<4;i++){a[i]=n%10;n=n/10;}
}//将数组里的数字转化为整数
int to_num(int a[]){int sum=0;for(int i=0;i<4;i++){sum=sum*10+a[i];}return sum;
}int main(int argc, const char * argv[]) {int n,MIN,MAX;scanf("%d",&n);int num[4];while(1){to_array(n,num);sort(num,num+4);MIN=to_num(num);sort(num,num+4,cmp);MAX=to_num(num);n=MAX-MIN;printf("%04d - %04d = %04d\n",MAX,MIN,n);//这里输入有格式要求，注意空格不能落if(n==0||n==6174)break;}return 0;
}