55D - Beautiful numbers

把lcm離散化一下就能過了。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long
using
 
 namespace std;

const int N = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;

int T, s[20], tot, hs[N], hcnt, who[N];
int Lcm[2521][2521];
LL L, R, f[20][2521][50];

inline int getId(int x) {
    return lower_bound(hs + 1
 
, hs + 1 + hcnt, x) - hs;
}

LL dp(int len, int mo, int lcm, bool limit) {
    if(len == -1) return mo % hs[lcm] ? 0 : 1;
    if(!limit && ~f[len][mo][lcm]) return f[len][mo][lcm];
    LL ans = 0;
    int up = limit ? s[len] : 9;
    for(int i = 0; i <= up; i++) {
        ans += dp(len - 1
 
, (mo*10+i)%2520, i ? who[Lcm[i][hs[lcm]]] : lcm, limit && up == i);
    }
    if(!limit) f[len][mo][lcm] = ans;
    return ans;
}

LL solve(LL x) {
    tot = 0;
    for(LL i = x; i; i /= 10) s[tot++] = i % 10;
    return dp(tot - 1, 0, 1, 1);
}

int main() {
    for(int i = 0; i < (1 << 9); i++) {
        int tmp = 1;
        for(int j = 0; j < 9; j++)
            if(i >> j & 1) tmp = tmp / __gcd(tmp, j + 1) * (j + 1);
        hs[++hcnt] = tmp;
    }
    sort(hs + 1, hs + 1 + hcnt);
    hcnt = unique(hs + 1, hs + 1 + hcnt) - hs - 1;
    for(int i = 1; i <= 2520; i++)
        who[i] = getId(i);
    memset(f, -1, sizeof(f));
    for(int i = 1; i <= 2520; i++)
        for(int j = 1; j <= 2520; j++)
            Lcm[i][j] = i / __gcd(i, j) * j;
    scanf("%d", &T);
    while(T--) {
        scanf("%lld%lld", &L, &R);
        printf("%lld\n", (solve(R) - solve(L - 1)));
    }
    return 0;
}

/*
*/

Codeforces 55D - Beautiful numbers 數位dp