bzoj1853: [Scoi2010]幸运数字

容斥搞一搞。。

不能直接1<<n枚举所有状态，得回溯。。从大的开始决定选不选，然后最好把能被其他数整除的数除掉。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>

using namespace std;

void setIO(const string& a) {
    freopen((a+".in").c_str(), "r", stdin);
    freopen((a+".out").c_str(), "w", stdout);
}

typedef long long ll;
const int N = 10;

const ll ten[11] = {1, (ll)1e1, (ll)1e2, (ll)1e3, (ll)1e4, (ll)1e5, (ll)1e6, (ll)1e7, (ll)1e8, (ll)1e9, (ll)1e10};

ll seq[2 << N], a[1 << N], res;
int tot;

void dfs_pre(int now, ll res) {
    if(res) seq[tot++] = res;
    if(now == N) return;
    dfs_pre(now + 1, res * 10 + 6);
    dfs_pre(now + 1, res * 10 + 8);
}

ll ans = 0;

ll gcd(ll a, ll b) {
    return !b ? a : gcd(b, a % b);
}

bool vis[1 << N];

int n;
ll L, R;

void dfs(int now, int cnt, ll lcm) {
    if(now == -1) {
        if(cnt & 1) ans += R / lcm - L / lcm;
        else if(cnt > 0) ans -= R / lcm - L / lcm;
        return;
    }
    dfs(now - 1, cnt, lcm);
    ll g = gcd(lcm, a[now]);
    if((double)lcm / g * a[now] <= R + 1) {
        dfs(now - 1, cnt + 1, lcm / g * a[now] );
    }
}

int main() {
    
    dfs_pre(0, 0);
    sort(seq, seq + tot);
    for(int i = 0; i < tot; i++) {
        int ok = 1;
        for(int j = 0; j < i; j++) if(seq[i] % seq[j] == 0) ok = 0;
        if(ok) a[n++] = seq[i];
    }
    
//    reverse(a, a + n);
    cin >> L >> R; --L;
    for(int i = n - 1; i >= 0; i--) {
        if(a[i] <= R) {
            dfs(i, 0, 1);
            break;
        }
    }
    cout << ans << endl;
    
    return 0;
}