PAT_A1069#The Black Hole of Numbers

Source:

PAT A1069 The Black Hole of Numbers (20 分)

Description:

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 (.

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

Keys:

• 字符串处理

Attention:

• 早期的PAT考试更注重算法的效率（过于依赖容器会超时），而现在的PAT考试更注重解决问题的能力（强调容器的使用）
• to_string()会超时

Code:

 #include<cstdio>
 #include<vector>
 #include<iostream>
 #include<algorithm>
 using namespace std;

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     int n1,n2,n;
     vector<int> p();
     scanf("%d",&n);
     do
     {
         for(int i=; i<; i++)
         {
             p[i]=n%;
             n/=;
         }
         sort(p.begin(),p.end(),less<char>());
         n1 = p[]*+p[]*+p[]*+p[];
         sort(p.begin(),p.end(),greater<char>());
         n2 = p[]*+p[]*+p[]*+p[];
         n = n2-n1;
         printf("%04d - %04d = %04d\n", n2,n1,n);
     }while(n!= && n!=);

     return ;
 }