[西安交大附中集训] 自积

Description

Solution

发现数字积数量很少, 枚举数字积 (S), 然后数位dp求数字积为 (S) 数的个数.

Code

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<set>
#include<cmath>
#include<cstdlib>
#include<cstring>
 
#define rep(i,l,r) for(int i=l;i<=r;++i)
#define repdo(i,l,r) for(int i=l;i>=r;--i)
 
using namespace std;
typedef long long ll;
typedef double db;
//----------------------------------------------
const int nsz=20;
ll l,r,bprod;
ll prod[10],v[10];
ll dp[nsz][35][25][15][15];
int add[10][10]={
   {0,0,0,0,0},
   {0,0,0,0,0},
   {0,1,0,0,0},
   {0,0,1,0,0},
   {0,2,0,0,0},
   {0,0,0,1,0},
   {0,1,1,0,0},
   {0,0,0,0,1},
   {0,3,0,0,0},
   {0,0,2,0,0}
};
 
ll dfs(int p,ll v,ll base,ll l,ll r){
    ll maxv=v+base-1;
    if(maxv<l||v>r)return 0;
    if(p==0)return !(prod[1]||prod[2]||prod[3]||prod[4]);
    ll &tmp=dp[p][prod[1]][prod[2]][prod[3]][prod[4]];
    if(l<=v&&maxv<=r&&~tmp)return tmp;
    ll res=0;
    base/=10;
    rep(i,(v!=0),9){
        bool ok=1;
        rep(j,1,4)ok&=(add[i][j]<=prod[j]);
        if(ok==0)continue;
        rep(j,1,4)prod[j]-=add[i][j];
        res+=dfs(p-1,v+i*base,base,l,r);
        rep(j,1,4)prod[j]+=add[i][j];
    }
    if(l<=v&&maxv<=r)tmp=res;
    return res;
}
 
ll sol(){
    bprod=sqrt(r);
    memset(dp,-1,sizeof(dp));
    ll res=0,tmp=1;
    v[1]=1;
    rep(i,0,30){
        if(i)v[1]*=2;
        v[2]=1;
        rep(j,0,20){
            if(j)v[2]*=3;
            if(v[1]*v[2]>bprod)break;
            v[3]=1; 
            rep(k,0,10){
                if(k)v[3]*=5;
                if(v[1]*v[2]*v[3]>bprod)break;
                v[4]=1;
                rep(a,0,10){
                    if(a)v[4]*=7;
                    ll tmp=v[1]*v[2]*v[3]*v[4];
                    if(tmp>bprod)break;
                    prod[1]=i,prod[2]=j,prod[3]=k,prod[4]=a;
                    res+=dfs(18,0,1e18,(l-1)/tmp+1,r/tmp);
                }
            }
        }
    }
    return res;
}
int main(){
    cin>>l>>r;
    cout<<sol()<<'
';
}