数论题 gcd 约数 dfs 容斥

 

 n<=15 a,b<=10^9 N<=10^4

#include<cstdio>
#define LL long long

int n, a, b, ans, t[20];

LL gcd(LL x, LL y) {
    if (!y) return x;
    return gcd(y, x%y);
}

void dfs(int cnt, int step, LL d) {
    if (d > b) return;
    if (step == n+1) {
        if (cnt&1) {
            ans -= b/d-(a-1)/d;
        } else {
            ans += b/d-(a-1)/d;
        }
        return;
    }
    dfs(cnt, step+1, d);
    dfs(cnt+1, step+1, d*t[step]/gcd(d, t[step]));
}

int main() {
    scanf("%d",&n);
    for (int i=1; i<=n;++i) {
        scanf("%d", &t[i]);
    }
    scanf("%d%d",&a,&b);
    dfs(0,1,6);
    printf("%d\n",ans);
    return 0;
}