How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6710    Accepted Submission(s): 1946

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 2
2 3

Sample Output

7

Author

wangye

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

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long Ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-10;
const int  inf =0x7f7f7f7f;
const double pi=acos(-1);
const int maxn=40000;

ll gcd(ll a,ll b)
{
    if(b==0) return a;
    else return gcd(b,a%b);
}

ll lcm(ll a,ll b)
{
    return (a/gcd(a,b))*b;
}

int n,m,bit,mm[24],tmp,cnt;
ll mult;
void solve(int flag)
{
    mult=1;bit=0;
    for(int i=0;i<cnt;i++)
       if(flag&(1<<i))
          {mult=lcm(mm[i],mult);bit++;}
}

int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        ll ans=0;n--;cnt=0;
        for(int i=0;i<m;i++)
        {
            scanf("%d",&tmp);
            if(tmp) mm[cnt++]=tmp;
        }
        for(int i=1;i<(1<<cnt);i++)
        {
            solve(i);
            int num=((ll)n)/mult;
            if(bit%2==1) ans+=num;
            else ans-=num;
        }
        printf("%lld\n",ans);
    }
    return 0;
}

分析：很好的一道容斥题，分析：求出在给定区间中能被集合中任意一个数整除的点的个数，分析题目的话
可以发现，先求出区间中所有能被集合中单个数整除的点的个数，求和后，会发现，能同时被两个数整除的点（是这两个数的最小公倍数的倍数）多算了一次，所以就减去能同时被两个数整除点的总个数，然后再加上能同时被三个点减去的点的个数.....(容斥)，不过这个题目有个很大的坑点，就是必须要去0，否则不仅会导致re，而且还会直接导致错误，因为在下面这段代码中，如果cnt换成m的话。可以发现0的存在就直接导致了ans的值得变化，所以必须要在读入集合时就直接将0剔除

for(int i=1;i<(1<<cnt);i++)
        {
            solve(i);
            int num=((ll)n)/mult;
            if(bit%2==1) ans+=num;
            else ans-=num;
        }