hdu 1796(容斥原理)

时间：2015-01-30 09:09:12 收藏：0 阅读：214

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1796

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4756 Accepted Submission(s): 1360

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

Author

wangye

题意：问在比n小的数里面，有多少个能被 M-integers set 中任意一个数整除的数？

思路：简单的容斥原理即可解决问题。用A1表示能被第1个数整除的数的个数，A2表示能被第2个数整除的数的个数，....Am表示能被第m个数整除的数的个数。

现在就是要求A1+A2+...Am -两两重复+三三重复-四四重复+五五重复.... （注意0得排掉，不算）

通过dfs暴力枚举重复数即可

#include <iostream>
#include <stdio.h>
using namespace std;
typedef long long ll;
ll a[110],ans,n,m;

int ct;

ll gcd(ll a, ll b)
{
   return b?gcd(b,a%b):a;
}

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

void dfs(int i,int cnt,ll sum,int pos)
{


   if(cnt==pos)
   {
     sum=(n-1)/sum;
     (pos&1)?ans+=sum:ans-=sum;
     return ;
   }
   if(i==ct+1)return;
   ll tmp=lcm(a[i],sum);
   dfs(i+1,cnt+1,tmp,pos);
   dfs(i+1,cnt,sum,pos);
}

void cal()
{
  ans=0;
  for(int i=1;i<=ct;i++)
  {
    dfs(1,0,1,i);
  }

}

int main()
{
  while(cin>>n>>m)
  {
    ct=0;
    for(int i=1;i<=m;i++)
    {
        ll tmp;
        scanf("%I64d",&tmp);
        if(!tmp)continue;
        a[++ct]=tmp;
    }
    cal();
    printf("%I64d\n",ans);
  }
  return 0;
}