找规律/数位DP HDOJ 4722 Good Numbers

 1 /*
 2     找规律/数位DP：我做的时候差一点做出来了，只是不知道最后的 is_one ()
 3                     http://www.cnblogs.com/crazyapple/p/3315436.html
 4     数位DP：http://blog.csdn.net/cyendra/article/details/11606209
 5 */
 6 #include <cstdio>
 7 #include <iostream>
 8 #include <algorithm>
 9 #include <cstring>
10 #include <string>
11 #include <cmath>
12 using namespace std;
13 
14 const int MAXN = 1e5 + 10;
15 const int INF = 0x3f3f3f3f;
16 
17 int is_one(long long n)
18 {
19     long long m = n;
20     long long i = n / 10 * 10;
21 
22     for (; i<=m; ++i)
23     {
24         long long tmp = i;    int sum = 0;
25         while (tmp)
26         {
27             sum += tmp % 10;
28             tmp /= 10;
29         }
30         if (sum % 10 == 0)    return 1;
31     }
32 
33     return 0;
34 }
35 
36 long long get_num(long long n)
37 {
38     if (n < 0)    return 0;
39     if (n <= 10)    return 1;
40 
41     return n/10 + is_one (n);
42 }
43 
44 int main(void)        //HDOJ 4722 Good Numbers
45 {
46     //freopen ("G.in", "r", stdin);
47 
48     int t, cas = 0;
49     long long l, r;
50     scanf ("%d", &t);
51     while (t--)
52     {
53         scanf ("%I64d%I64d", &l, &r);
54 
55         printf ("Case #%d: %I64d\n", ++cas, get_num (r) - get_num (l-1));
56     }
57 
58     return 0;
59 }
60 
61 
62 /*
63 Case #1: 0
64 Case #2: 1
65 Case #3: 1
66 */