pat 1069. The Black Hole of Numbers (20)

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:

Sample Output 1:

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

Sample Input 2:

Sample Output 2:

2222 - 2222 = 0000

解：题目有点小坑，我看题目描述以为它应该指定输入四位数，虽然它写的是(0,10000)，但是题目也有一句话：Given any 4-digit number, you are supposed to illustrate the way it gets into the black hole.总之需要考虑当输入的数值为1位2位3位的情况。

代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int main()
{
    int cnt=0,n,a[5],temp,num1,num2;
    scanf("%d",&n);
    int hhd=n;
    do
    {
        cnt=0;
        while(hhd)
        {
            a[cnt++]=hhd%10;
            hhd=hhd/10;
        }
        while(cnt<4)
        {
            a[cnt++]=0;
        }
        sort(a,a+cnt);
        if(a[0]==a[1]&&a[1]==a[2]&&a[2]==a[3])
        {
            printf("%d - %d = 0000",n,n);
            break;
        }
        num1=a[3]*1000+a[2]*100+a[1]*10+a[0];
        num2=a[0]*1000+a[1]*100+a[2]*10+a[3];
        cout<<a[3]<<a[2]<<a[1]<<a[0]<<" - "<<a[0]<<a[1]<<a[2]<<a[3]<<" = ";
        hhd=num1-num2;
        if(hhd>=1000&&hhd<=9999)
            cout<<num1-num2<<endl;
        else
        {
            if(hhd>=100&&hhd<=999)
                cout<<'0'<<hhd<<endl;
            if(hhd>=10&&hhd<=99)
                cout<<'00'<<hhd<<endl;
            if(hhd>=1&&hhd<=9)
                cout<<'000'<<hhd<<endl;
        }
    }while(hhd!=6174);
}