題意：給定一個數 n，和一個集合 m，問你小於的 n的所有正數能整除 m的任意一個的數目。

析：簡單容斥，就是 1 個數的倍數 - 2個數的最小公倍數 + 3個數的最小公倍數 + ...(-1)^(n+1) * n個數的最小公倍數。

代碼如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

int a[maxn];
int lcm(int a, int b){
  return a * (b / gcd(a, b));
}

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    for(int i = 0; i < m; ++i)  scanf("%d", a+i);
    int all = 1<<m;
    int ans = 0;
    --n;
    for(int i = 1; i < all; ++i){
      int cnt = 0, l = 1;
      for(int j = 0; j < m; ++j) if(i&(1<<j)){
        ++cnt;
        l = lcm(l, a[j]);
      }
      if(l == 0)  continue;
      ans += (cnt&1) ? n / l : - n / l;
    }

    printf("%d\n", ans);
  }
  return 0;
}

HDU 1796 How many integers can you find (容斥)