hdu1796How many integers can you find

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5795    Accepted Submission(s): 1668

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

 

Author

wangye

 

Source

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

/*题目大意：给定n和一个大小为m的集合，集合元素为非负整数。为1...n内能被集合里任意一个数整除的数字个数。n<=2^31,m<=10解题思路：容斥原理地简单应用。先找出1...n内能被集合中任意一个元素整除的个数，再减去能被集合中任意两个整除的个数，即能被它们两只的最小公倍数整除的个数，因为这部分被计算了两次，然后又加上三个时候的个数，然后又减去四个时候的倍数...最后判断下集合元素的个数为奇还是偶，奇加偶减。*/#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <ctime>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define inf -0x3f3f3f3f#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define mem0(a) memset(a,0,sizeof(a))#define mem1(a) memset(a,-1,sizeof(a))#define mem(a, b) memset(a, b, sizeof(a))typedef long long ll;int n,m;vector<int>p;int gcd(int a,int b){    if(a<b)        swap(a,b);    if(b==0)        return a;    return gcd(b,a%b);}int main(){    while(scanf("%d%d",&n,&m)!=EOF){        p.clear();        int x;        for(int i=1;i<=m;i++){            scanf("%d",&x);            if(x!=0)                p.push_back (x);        }        int sum=0;        for(int msk=1;msk<(1<<p.size());msk++){            int mult=1,bits=0;            for(int i=0;i<(int)p.size();i++)                if(msk&(1<<i)){                    ++bits;                    mult=p[i]/gcd(p[i],mult)*mult;  //可能爆int                }            int cur=(n-1)/mult;            if(bits&1)                sum+=cur;            else                sum-=cur;        }        printf("%d\n",sum);    }    return 0;}