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:

 1 #include<cstdio>
 2 #include<vector>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 
 7 int main()
 8 {
 9 #ifdef ONLINE_JUDGE
10 #else
11     freopen("Test.txt", "r", stdin);
12 #endif // ONLINE_JUDGE
13 
14     int n1,n2,n;
15     vector<int> p(4);
16     scanf("%d",&n);
17     do
18     {
19         for(int i=0; i<4; i++)
20         {
21             p[i]=n%10;
22             n/=10;
23         }
24         sort(p.begin(),p.end(),less<char>());
25         n1 = p[0]*1000+p[1]*100+p[2]*10+p[3];
26         sort(p.begin(),p.end(),greater<char>());
27         n2 = p[0]*1000+p[1]*100+p[2]*10+p[3];
28         n = n2-n1;
29         printf("%04d - %04d = %04d\n", n2,n1,n);
30     }while(n!=0 && n!=6174);
31 
32     return 0;
33 }

 

posted @ 2019-08-30 15:04  林東雨  阅读(1252)  评论(0)  编辑  收藏  举报