【BZOJ1183】[Croatian2008]Umnozak【数位dp】

考虑到$n$的$digit-product$一定是$2,3,5,7$的倍数，且不超过$\sqrt{n}$，所以可以枚举$digit-product$，然后记忆化搜索就好了。。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define int long long
using namespace std;
int l,r,tot,cc,num[20],f[20][40][30][20][20],add[10][4]={{0,0,0,0},{0,0,0,0},{1,0,0,0},{0,1,0,0},{2,0,0,0},{0,0,1,0},{1,1,0,0},{0,0,0,1},{3,0,0,0},{0,2,0,0}};
bool vis[20][40][30][20][20];
int dfs(int now,int ta,int tb,int tc,int td,bool flag,bool isfir){
    if(now==0){
        return (ta+tb+tc+td)==0&&!isfir;
    }
    if(!flag&&!isfir&&vis[now][ta][tb][tc][td]){
        return f[now][ta][tb][tc][td];
    }
    int res=0;
    if(isfir){
        res+=dfs(now-1,ta,tb,tc,td,flag&&num[now]==0,true);
    }
    for(int i=1,up=flag?num[now]:9;i<=up;i++){
        if(ta-add[i][0]>=0&&tb-add[i][1]>=0&&tc-add[i][2]>=0&&td-add[i][3]>=0){
            res+=dfs(now-1,ta-add[i][0],tb-add[i][1],tc-add[i][2],td-add[i][3],flag&&i==num[now],false);
        }
    }
    if(!flag&&!isfir){
        vis[now][ta][tb][tc][td]=true;
        f[now][ta][tb][tc][td]=res;
    }
    return res;
}
int calc(int x){
    int tmp,res=0;
    for(int a=1,ta=0;a*a<=x;a*=2,ta++){
        for(int b=a,tb=0;b*b<=x;b*=3,tb++){
            for(int c=b,tc=0;c*c<=x;c*=5,tc++){
                for(int d=c,td=0;d*d<=x;d*=7,td++){
                    num[0]=0;
                    tmp=x/d;
                    do{
                        num[++num[0]]=tmp%10;
                        tmp/=10;
                    }while(tmp);
                    res+=dfs(num[0],ta,tb,tc,td,true,true);
                }
            }
        }
    }
    return res;
}
signed main(){
    scanf("%lld%lld",&l,&r);
    printf("%lld\n",calc(r)-calc(l-1));
    return 0;
}