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/libin56842/article/details/11580497
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
", ++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 */
1 /*
2 数位DP：基础题，对于一个数，把它每位数分出来，从最高位开始枚举。dp[i][(j+k)%10] += dp[i+1][j];
3 dp[i][j]表示前i位数和取模是j的个数
4 */
5 #include <cstdio>
6 #include <algorithm>
7 #include <cstring>
8 #include <cmath>
9 using namespace std;
10
11 typedef long long ll;
12 const int MAXN = 22;
13 const int INF = 0x3f3f3f3f;
14 ll dp[MAXN][MAXN];
15 int dig[MAXN];
16
17 ll work(ll n) {
18 ll tmp = n; int len = 0;
19 memset (dig, 0, sizeof (dig));
20 while (tmp) {
21 dig[++len] = tmp % 10;
22 tmp /= 10;
23 }
24 memset (dp, 0, sizeof (dp)); int x = 0;
25 for (int i=len; i>=1; --i) {
26 for (int j=0; j<10; ++j) {
27 for (int k=0; k<10; ++k) {
28 dp[i][(j+k)%10] += dp[i+1][j];
29 }
30 }
31 for (int j=0; j<dig[i]; ++j) dp[i][(x+j)%10]++;
32 x = (x + dig[i]) % 10;
33 }
34 if (!x) dp[1][0]++;
35 return dp[1][0];
36 }
37
38 int main(void) { //HDOJ 4722 Good Numbers
39 int T, cas = 0; scanf ("%d", &T);
40 while (T--) {
41 ll l, r; scanf ("%I64d%I64d", &l, &r);
42 printf ("Case #%d: %I64d
", ++cas, work (r) - work (l-1));
43 }
44
45 return 0;
46 }