codeforces55D数位dp

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <stdlib.h>
 4 #include <algorithm>
 5 #include <iostream>
 6 #include <queue>
 7 #include <stack>
 8 #include <vector>
 9 #include <map>
10 #include <set>
11 #include <string>
12 #include <math.h>
13 using namespace std;
14 #pragma warning(disable:4996)
15 #pragma comment(linker, "/STACK:1024000000,1024000000")
16 typedef __int64 LL;      
17 const int mod = 2520;
18 const int INF = 1<<30;
19 /*
20 dp[i][j][k]
21 表示前取了前i位，数字为j，各个位的最小公倍数位k
22 如果j%k==0  那么说明可以
23 
24 dp[i+1][j*10 + i][lcm(k,i)] = dp[i][j][k]
25 
26 */
27 
28 LL dp[30][2520][50];
29 int num[64];
30 bool vis[10000];
31 int hs[2520 + 10];
32 int gcd(int a, int b)
33 {
34     if (b == 0)return a;
35     return gcd(b, a%b);
36 }
37 int lcm(int a, int b)
38 {
39     return a*b / gcd(a, b);
40 }
41 void init()
42 {
43     int cnt = 0;
44     for (int i = 1; i <= mod; ++i)
45     if (mod%i == 0)
46         hs[i] = cnt++;
47 }
48 
49 LL dfs(int pos, int preSum, int preLcm, bool flag)
50 {
51     if (pos == 0)return preSum % preLcm==0;
52     if (!flag && dp[pos][preSum][hs[preLcm]] != -1)
53         return dp[pos][preSum][hs[preLcm]];
54     int limit = flag ? num[pos] : 9;
55     LL ans = 0;
56     for (int i = 0; i <= limit; ++i)
57     {
58         int nowSum = (preSum * 10 + i) % mod;
59         int nowLcm = preLcm;
60         if (i)nowLcm = lcm(nowLcm, i);
61         ans += dfs(pos - 1, nowSum, nowLcm, flag&&i == limit);
62     }
63     if (!flag)//如果pos为比n的第pos位小，那么说明可以利用已有的结果，如果不是的话，不能，因为可能比n大
64         dp[pos][preSum][hs[preLcm]] = ans;
65     return ans;
66 }
67 LL calc(LL n)
68 {
69 
70     int len = 0;
71     while (n)
72     {
73         num[++len] = n % 10;
74         n /= 10;
75     }
76     
77     return dfs(len,0,1,1);
78 }
79 int main()
80 {
81     init();
82     int t;
83     LL l, r;
84     memset(dp, -1, sizeof(dp));//只初始化一次，因为计算的结果可多次利用
85     scanf("%d", &t);
86     while (t--)
87     {
88         scanf("%I64d%I64d", &l, &r);
89         printf("%I64d\n", calc(r) - calc(l-1));
90     }
91     return 0;
92 }