HDU 1796 How many integers can you find (容斥原理)

题目链接：How many integers can you find

题面：

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5345    Accepted Submission(s): 1515

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

 

Source

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

 

Recommend

wangye

解题：

     容斥原理是早学过，但从没想过功能如此强大，最近老是碰到。这道题，题意是求给定的集合中，在1~（n-1）范围内，能找到多少个数，为集合中任一数的倍数。其实也就是求这些数倍数集合的并。以两个为例，根据容斥原理，结果等于集合A∪集合B-集合A∩B。A∩B，即为既是A的倍数，又是B的倍数，那么就是A,B最小公倍数的倍数。模拟二进制表示，某位取或不取，最后容斥一下，就得到结果了。神坑的是，很容易看错，是会有0出现的，需要排除掉。

代码：

#include <iostream>
#include <cstring>
using namespace std;
int one_amount[1050];
int refl[1050][10];
void cal()
{
    memset(refl,0,sizeof(refl));
    int cnt=0,temp;
   for(int i=0;i<1024;i++)
   {
       temp=i;
       one_amount[i]=cnt=0;
       while(temp)
       {
         if(temp%2)
         {
             one_amount[i]++;
             refl[i][cnt]=1;
         }
         cnt++;
         temp/=2;
       }
   }
}
long long gcd(long long a,long long b)
{
    if(a==0)return b;
    else return gcd(b%a,a);
}
int main()
{
    int m,x;
    unsigned long long n,store[12],tmp,ans,temp;
    cal();
    while(cin>>n>>m)
    {
        ans=0;
        int y=m,z=0;
       while(y--)
       {
           cin>>store[z];
           if(store[z]==0)
           {
               m--;
               continue;
           }
           else z++;
        }
       x=1;
       for(int i=0;i<m;i++)
           x*=2;
       for(int i=1;i<x;i++)
       {
            tmp=1;
            for(int j=0;j<m;j++)
            {
                if(refl[i][j])
                {
                    temp=gcd(tmp,store[j]);
                    tmp=tmp*store[j]/temp;
                }
            }
            if(one_amount[i]%2)
            ans+=(n-1)/tmp;
            else
            ans-=(n-1)/tmp;
       }
       cout<<ans<<endl;
    }
    return 0;
}