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

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
",sum);
    }
    return 0;
}

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

//968MS