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): 8059    Accepted Submission(s): 2403


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

 

题意:给你一个n和一个m个数的集合，求1到n中满足能被集合中一个数整除的数有几个

思路:根据容斥定理，满足条件的数为被集合中一个数整除的数的数量和-能被集合中任意两个数整除的数的数量+三个......

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

typedef long long ll;

int a[20];
int n,m;

void solve()
{
int res=0;
for(int i=1;i<(1<<m);i++)
{
int num=0;
for(int j=i;j!=0;j>>=1) num+=j&1;
int lcm=1;
for(int j=0;j<m;j++)
{
if(i>>j&1)
{
lcm=lcm/__gcd(lcm,a[j])*a[j];
if(lcm>n) break;
}
}
if(num%2==0) res-=(n/lcm);
else res+=(n/lcm);
}
printf("%d\n",res);
}

int main()
{

while(~scanf("%d%d",&n,&m))
{
n--;
int flag=1;
for(int i=0;i<m;i++)
{
scanf("%d",&a[i]);
if(a[i]==0)
{
i--;m--;
}
}
solve();
}
return 0;
}