pat乙级1019

求四位数的每一位：

1 int a[4];
2 for (int i = 0; i < 4; i++)
3 {
4    a[i] = n % 10;
5    n /= 10;
6 }

输出4位整数，少于4位，高位补0：

1 printf("%04d", n);

6174是一个测试点，所以用do....while语句。

 1 #include <iostream>
 2 #include <string>
 3 
 4 using namespace std;
 5 
 6 int up, down;
 7 void getNumber(int n)
 8 {
 9     int a[4];
10     for (int i = 0; i < 4; i++)
11     {
12         a[i] = n % 10;
13         n /= 10;
14     }
15     for (int i = 0; i < 4; i++)
16     {
17         for (int j = i; j > 0 && a[j] > a[j - 1]; j--)
18         {
19             int temp = a[j];
20             a[j] = a[j - 1];
21             a[j - 1] = temp;
22         }
23     }
24     
25     down = a[0] * 1000 + a[1] * 100 + a[2] * 10 + a[3];
26     up = a[3] * 1000 + a[2] * 100 + a[1] * 10 + a[0];
27 }
28 
29 int main()
30 {
31     
32     int n;
33     cin >> n;
34 
35     do {
36         getNumber(n);
37         n = down - up;
38         printf("%04d - %04d = %04d\n", down, up, n);
39     } while (n != 6174 && n != 0000);
40 
41     return 0;
42 }

 

posted @ 2018-02-22 12:10  bloglxc  阅读(112)  评论(0)    收藏  举报