Source:
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 until6174comes 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 }