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。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;
 #define LL __int64
 #define MOD 2520
 LL dp[][][];
 int index[];
 int num[];
 LL gcd(LL a,LL b)
 {
     return b == ?a:gcd(b,a%b);
 }
 LL lcm(LL a,LL b)
 {
     return a*b/gcd(a,b);
 }
 LL dfs(int pos,int pre,int mod,int bound)
 {
     int end,tpre,tmod,i;
     LL ans = ;
     if(pos == -)
     return pre%mod == ;
     if(!bound&&dp[pos][pre][index[mod]] != -)
     return dp[pos][pre][index[mod]];
     end = bound ? num[pos] : ;
     for(i = ;i <= end;i ++)
     {
         tpre = (pre* + i)%MOD;
         if(i)
         tmod = lcm(mod,i);
         else
         tmod = mod;
         ans += dfs(pos-,tpre,tmod,bound&&i == end);
     }
     if(!bound)
     dp[pos][pre][index[mod]] = ans;
     return ans;
 }
 LL judge(LL x)
 {
     int pos = ;
     while(x)
     {
         num[pos++] = x%;
         x = x/;
     }
     return dfs(pos-,,,);
 }
 int main()
 {
     int t,i,num = ;
     LL x,y;
     memset(dp,-,sizeof(dp));
     for(i = ;i <= MOD;i ++)
     {
         if(MOD%i == )
         index[i] = num++;
     }
     cin>>t;
     while(t--)
     {
         cin>>x>>y;
         printf("%I64d\n",judge(y)-judge(x-));
     }
     return ;
 }