UVaOJ1225 Digit Counting AC代码示例

 1 #include<iostream>
 2 #include<string.h>
 3 #include<math.h>
 4 using namespace std;
 5 
 6 int s[10009][11];
 7 
 8 int main()
 9 {
10     memset(s, 0, sizeof(s));
11     for (int i = 1; i < 10009; i++) {
12         s[i][0] = i;
13         s[i][1] = s[i - 1][1];
14         s[i][2] = s[i - 1][2];
15         s[i][3] = s[i - 1][3];
16         s[i][4] = s[i - 1][4];
17         s[i][5] = s[i - 1][5];
18         s[i][6] = s[i - 1][6];
19         s[i][7] = s[i - 1][7];
20         s[i][8] = s[i - 1][8];
21         s[i][9] = s[i - 1][9];
22         s[i][10] = s[i - 1][10];
23 
24         int n = s[i][0];
25         while (n) {
26             int m = n % 10;
27             switch (m) {
28             case 0:        s[i][1] +=  1;    break;
29             case 1:        s[i][2] +=  1;    break;
30             case 2:        s[i][3] += 1;    break;
31             case 3:        s[i][4] +=  1;    break;
32             case 4:        s[i][5] +=  1;    break;
33             case 5:        s[i][6] +=  1;    break;
34             case 6:        s[i][7] +=  1;    break;
35             case 7:        s[i][8] +=  1;    break;
36             case 8:        s[i][9] +=  1;    break;
37             case 9:        s[i][10]+=  1;    break;
38             }
39             n /= 10;
40         }
41     }
42 
43     int T = 0;
44     cin >> T;
45     while (T--) {
46         int num;
47         cin >> num;
48         printf("%d %d %d %d %d %d %d %d %d %d\n",
49             s[num][1], s[num][2], s[num][3], s[num][4], s[num][5], s[num][6], s[num][7], s[num][8], s[num][9], s[num][10]);
50     }
51     return 0;
52 }