HDU 1796 How many integers can you find (lcm + 容斥)

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 5526    Accepted Submission(s): 1584

Problem Description

  Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10}, all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7.

 

Input

  There are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.

 

Output

  For each case, output the number.

 

Sample Input

12 22 3

 

Sample Output

7

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1796

题目大意：给一个数n和一个含有m个数的集合，求[1, n-1]中可以被集合中任意一个数整除的数集大小

题目分析：还算比较裸的容斥了，有一个大坑点，集合中可能有0，如果有0的话，直接把0去掉，还可以加个小特判，如果有1的话直接输出n就行了，注意这里容斥不是把数字直接相乘，因为它们不一定互质，所以算的是数字间的最小公倍数，举个简单例子：
25 2
6 8
答案如果直接用24 / 6 + 24 / 8 - 24 / (6 * 8)得到的是7，显然答案是6，因为24没有被容斥掉，所以我们要拿两个数的最小公倍数去容斥

#include <cstdio>#include <cstring>#include <algorithm>#define ll long longusing namespace std;int a[25], b[25];ll ans;int n, m, cnt;ll gcd(ll a, ll b){    return b ? gcd(b, a % b) : a;  }ll lcm(int a, int b){    return (ll) a * b / gcd(a, b);}void DFS(int idx, ll cur, int sgin){    for(int i = idx; i < cnt; i++)    {        ll tmp = lcm(cur, (ll)b[i]);        ans += (ll) sgin * (n / tmp);        DFS(i + 1, tmp, -sgin);    }}int main(){    while(scanf("%d %d", &n, &m) != EOF)    {        n --;        cnt = 0;        bool flag = false;        for(int i = 0; i < m; i++)        {            scanf("%d", &a[i]);            if(a[i] != 0)                b[cnt ++] = a[i];            if(a[i] == 1)                flag = true;        }        if(flag)        {            printf("%d\n", n);            continue;        }        sort(b, b + cnt);        ans = 0;        DFS(0, 1, 1);        printf("%I64d\n", ans);    }}