HDU 1796How many integers can you find(简单容斥定理)

 

HDU 1796How many integers can you find(简单容斥定理)

分类： 数论 2013-10-12 20:59  98人阅读  评论(0)  收藏  举报

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): 3315    Accepted Submission(s): 937

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）

 

题目大意：很简单的题目，直接看意思就懂哈！

      解题思路：容斥定理，加奇减偶，开始忘记求lcm了，囧！！而且开始还特判0的情况，题目中说的必须是除以，所以0不是一个解。。。开始竟然以为需要是因子就可以了。想通了之后直接先筛选一次，把0都筛选出去。

      题目地址：How many integers can you find

AC代码：

 #include<iostream>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<cstdio>
 using namespace std;
 __int64 sum;
 int n,m;
 int a[25];
 int b[25];
 int visi[25];

 __int64 gcd(__int64 m,__int64 n)
 {
     __int64 tmp;
     while(n)
     {
         tmp=m%n;
         m=n;
         n=tmp;
     }
     return m;
 }

 __int64 lcm(__int64 m,__int64 n)
 {
     return m/gcd(m,n)*n;
 }

 void cal()
 {
     int flag=0,i;
     __int64 t=1;
     __int64 ans;
     for(i=0;i<m;i++)
     {
         if(visi[i])
         {
             flag++;   //记录用了多少个数
             t=lcm(t,b[i]);
         }
     }
     ans=n/t;
     if(n%t==0) ans--;
     if(flag&1) sum+=ans;   //加奇减偶
     else sum-=ans;
 }

 int main()
 {
     int i,j,p;
     while(~scanf("%d%d",&n,&m))
     {
         sum=0;
         for(i=0;i<m;i++)
             scanf("%d",&a[i]);

         int tt=0;  //
         for(i=0;i<m;i++)
         {
             if(a[i])  //去掉0
                 b[tt++]=a[i];
         }
         m=tt;
         p=1<<m;   //p表示选取多少个数，组合数的状态
         for(i=1;i<p;i++)
         {
             int tmp=i;
             for(j=0;j<m;j++)
             {
                 visi[j]=tmp&1;
                 tmp>>=1;
             }
             cal();
         }
         printf("%I64d\n",sum);
     }
     return 0;
 }

 /*
 12 2
 2 3
 12 3
 2 3 0
 12 4
 2 3 2 0
 */

 //968MS