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

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7456    Accepted Submission(s): 2205


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

题意：给定n和一个大小为m的集合，集合元素为非负整数。求1...n内能被集合里任意一个数整除的数字个数。

 
思路：容斥原理的简单应用。

先找出1...n内能被集合中任意一个元素整除的个数，再减去能被集合中任意两个整除的个数，即能被它们两只的最小公倍数整除的个数，因为这部分被计算了两次，然后又加上三个的时候的个数，然后又减去四个时候的倍数...判断集合元素的个数为奇还是偶，奇加偶减。

注意：集合中的数可能有0存在，此时要排除0。

AC代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<vector>
typedef __int64 LL;
using namespace std;
LL a[105];

LL gcd(LL a,LL b)
{
return b==0?a:gcd(b,a%b);
}
LL solve(LL n,LL m)
{
LL ans=0;
for(LL msk=1; msk<(1<<m); msk++) //共2^m种组合情况
{
LL mult=1,cnt=0;
for(LL i=0; i<m; i++)
{
if(msk&(1<<i))
{
mult=mult/gcd(mult,a[i])*a[i]; //计算当前组合的最小公倍数
++cnt;
}
}
LL ks=(n-1)/mult;
if(cnt&1)ans+=ks;
else ans-=ks;
}
return ans;
}

int main()
{
LL n,m;
while(~scanf("%I64d%I64d",&n,&m))
{
for(int i=0; i<m; i++)
{
scanf("%I64d",a+i);
if(a[i]==0)i--,m--;
}
printf("%I64d\n",solve(n,m));
}
return 0;
}