[ZOJ 2836] Number Puzzle

Number Puzzle

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Given a list of integers (A₁, A₂, ..., A_n), and a positive integer M, please find the number of positive integers that are not greater than M and dividable by any integer from the given list.

Input

The input contains several test cases.

For each test case, there are two lines. The first line contains N (1 <= N <= 10) and M (1 <= M <= 200000000), and the second line contains A₁, A₂, ..., A_n(1 <= A_i <= 10, for i = 1, 2, ..., N).

Output

For each test case in the input, output the result in a single line.

Sample Input

3 2 2 3 7 3 6 2 3 7

Sample Output

1 4

入门容斥原理，见代码

#include <iostream>

#include <cstdio>

using namespace std;

#define ll long long

#define N 10



int n,m;

int a[N];

int gcd(ll a,ll b){

    return b?gcd(b,a%b):a;

}

int lcm(ll a,ll b){

    return a/gcd(a,b)*b;

}

////写法1/////

int solve()

{

    int res=0;

    for(int i=1;i<(1<<n);i++)

    {

        int cnt=0;

        ll LCM=1;

        for(int j=0;j<n;j++)

        {

            if(i&(1<<j)) cnt++,LCM=lcm(LCM,a[j]);

        }

        if(cnt&1) res+=m/LCM;

        else res-=m/LCM;

    }

    return res;

}

////写法2/////

int ans;

void dfs(int pos,int LCM,int cnt)

{

    if(cnt)

    {

        if(cnt&1) ans+=m/LCM;

        else ans-=m/LCM;

    }

    for(int i=pos+1;i<n;i++)

        dfs(i,lcm(LCM,a[i]),cnt+1);

}

int main()

{

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        for(int i=0;i<n;i++) scanf("%d",&a[i]);

        //printf("%d\n",solve());

        ans=0;

        dfs(-1,1,0);

        printf("%d\n",ans);

    }

    return 0;

}