CF 55D. Beautiful numbers(数位DP)

题目链接

这题，没想出来，根本没想到用最小公倍数来更新，一直想状态压缩，不过余数什么的根本存不下，看的von学长的blog，比着写了写，就是模版改改，不过状态转移构造不出，怎么着，都做不出来。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 using namespace std;

 5 #define LL __int64

 6 #define MOD 2520

 7 LL dp[31][2600][50];

 8 int index[3000];

 9 int num[31];

10 LL gcd(LL a,LL b)

11 {

12     return b == 0?a:gcd(b,a%b);

13 }

14 LL lcm(LL a,LL b)

15 {

16     return a*b/gcd(a,b);

17 }

18 LL dfs(int pos,int pre,int mod,int bound)

19 {

20     int end,tpre,tmod,i;

21     LL ans = 0;

22     if(pos == -1)

23     return pre%mod == 0;

24     if(!bound&&dp[pos][pre][index[mod]] != -1)

25     return dp[pos][pre][index[mod]];

26     end = bound ? num[pos] : 9;

27     for(i = 0;i <= end;i ++)

28     {

29         tpre = (pre*10 + i)%MOD;

30         if(i)

31         tmod = lcm(mod,i);

32         else

33         tmod = mod;

34         ans += dfs(pos-1,tpre,tmod,bound&&i == end);

35     }

36     if(!bound)

37     dp[pos][pre][index[mod]] = ans;

38     return ans;

39 }

40 LL judge(LL x)

41 {

42     int pos = 0;

43     while(x)

44     {

45         num[pos++] = x%10;

46         x = x/10;

47     }

48     return dfs(pos-1,0,1,1);

49 }

50 int main()

51 {

52     int t,i,num = 0;

53     LL x,y;

54     memset(dp,-1,sizeof(dp));

55     for(i = 1;i <= MOD;i ++)

56     {

57         if(MOD%i == 0)

58         index[i] = num++;

59     }

60     cin>>t;

61     while(t--)

62     {

63         cin>>x>>y;

64         printf("%I64d\n",judge(y)-judge(x-1));

65     }

66     return 0;

67 }