BZOJ3679: 数字之积(数位dp)

题意

题目链接

Sol

推什么结论啊。

直接大力dp，$f[i][j]$表示第$i$位，乘积为$j$，第二维直接开map

能赢！

/*

*/
#include<iostream>
#include<cstdio>
#include<map>
#define LL long long
using namespace std;
inline LL read() {
    char c = getchar(); LL x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
LL L, R, L1, R1;
map<LL, LL> f[];
LL s[], num = ;
LL dfs(LL x, bool lim, LL mul) {
    if(x < ) return ;
    if(x == ) {
        if(mul == -) mul = ;
        return mul >= L1 && mul <= R1;
    }
    if((!lim) && (f[x].find(mul) != f[x].end())) return f[x][mul];
    LL ans = ;
    for(int i = ; i <= (lim ? s[x] : ); i++) {
        if(mul == -) {
            if(i == ) ans += dfs(x - , lim && (i == s[x]), -);
            else ans += dfs(x - , lim && (i == s[x]), i);
        } else ans += dfs(x - , lim && (i == s[x]), i * mul);
    }
    if(!lim) f[x][mul] = ans;
    return ans;
}
LL solve(LL x) {
    if(x == -) return ;
    num = ;
    while(x) s[++num] = x % , x /= ;
    return dfs(num, , -);
}
int main() {
    R1 = read();
    L = read(); R = read() - ; L1 = ;
    if(L == R + ) {
        printf("%lld", L >  && L <= R1); return ;
    }
    LL ans = solve(R) - solve(L - );
    cout << ans;
    return ;
}
/*
23333
123456789 123456789123456789

6000000
123456 12345678

6
100 113

6
0 3
*/