【HDU】1796 How many integers can you find

题意：给出n和m个元素的集合，求小于n且是集合中任意元素的倍数的数有多少个(m<=10)。

最基础的容斥了，懒得DFS，直接枚举二进制暴力完事。

#include<cstdio>
#include<algorithm>
typedef long long LL;
#define MAXN 110
using namespace std;
int n, m, ans;
int a[MAXN];
LL GCD(LL x, LL y) {
    return y ? GCD(y, x % y) : x;
}
LL LCM(LL x, LL y) {
    return x / GCD(x, y) * y;
}
LL Get(int x, int &cnt) {
    int i;
    LL res = 1;
    for (i = cnt = 0; i < m; i++) {
        if (x & (1 << i)) {
            cnt++;
            res = LCM(res, a[i]);
        }
    }
    return res;
}
int main() {
    int i, k;
    LL tmp;
    while (~scanf("%d%d", &n, &m)) {
        n--;
        k = m;
        for (ans = i = 0; i < m; i++) {
            scanf("%d", &a[i]);
            if (a[i] == 0)
                k = i;
        }
        if (k != m) {
            m--;
            swap(a[m], a[k]);
        }
        for (i = 1; i < (1 << m); i++) {
            tmp = Get(i, k);
            if (k & 1)
                ans += n / tmp;
            else
                ans -= n / tmp;
        }
        printf("%d\n", ans);
    }
    return 0;
}