1069. The Black Hole of Numbers

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 = 10899810 - 0189 = 96219621 - 1269 = 83528532 - 2358 = 6174

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000

之前写过这个题，看了一下，还可以，就不改了，按流程来即可。

#include<iostream>#include<string>#include<algorithm>#include<iomanip>using namespace std;int main(){int n,s=0;int a[4],b[4]={0,0,0,0};cin>>n;while(s!=6174){for(int i=0;i<4;i++){a[i]=n%10;n=n/10;}if(a[0]==a[1]&&a[0]==a[2]&&a[0]==a[3]){cout<<a[0]<<a[1]<<a[2]<<a[3]<<" - ";cout<<a[0]<<a[1]<<a[2]<<a[3]<<" = 0000"<<endl;break;}sort(a,a+4);for(int i=0;i<4;i++)b[i]=a[3-i];int a1=0,b1=0;for(int i=0;i<4;i++){a1=a1*10+a[i];b1=b1*10+b[i];}s=b1-a1;n=s;cout<<setfill('0')<<setw(4)<<b1<<" - "<<setfill('0')<<setw(4)<<a1<<" = "<<setfill('0')<<setw(4)<<s<<endl;}return 0;}