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[编程语言] UVA 10325 The Lottery(容斥)
UVA 10325 The Lottery(容斥)


 The Lottery 


The Sports Association of Bangladesh is in great problem with their latest lottery 'Jodi laiga Jai'. There are so many participants this time that they cannot manage all the numbers. In an urgent meeting they have decided that they will ignore some numbers. But how they will choose those unlucky numbers!! Mr. NondoDulal who is very interested about historic problems proposed a scheme to get free from this problem.
You may be interested to know how he has got this scheme. Recently he has read the Joseph's problem.

The Problem

There are N tickets which are numbered from 1 to N. Mr. Nondo will choose M random numbers and then he will select those numbers which is divisible by at least one of those M numbers. The numbers which are not divisible by any of those M numbers will be considered for the lottery.
As you know each number is divisible by 1. So Mr. Nondo will never select 1 as one of those M numbers. Now given N,M and M random numbers, you have to find out the number of tickets which will be considered for the lottery.

The Input


Each input set starts with two Integers N (10<=N<2^31) and M (1<=M<=15). The next line will contain M positive integers each of which is not greater than N. Input is terminated by EOF.

The Output


Just print in a line out of N tickets how many will be considered for the lottery.

Sample Input


10 2
2 3
20 2
2 4

Sample Output


3
10

Md. Kamruzzaman


把出局的人找出来拿n减去出局的人就可以了。
考虑容斥做法。设f[i]表示i个数的最小公倍数。那么出局总人数s=f[1]-f[2]+f[3]..
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int MOD=1e9+7;
typedef long long LL;
vector<int>v;
LL n,m;
LL a[20];
LL gcd(LL a,LL b)
{
    return b?gcd(b,a%b):a;
}
LL lcm(LL a,LL b)
{
    return a/gcd(a,b)*b;
}
void solve()
{
    LL ans=0,lc;
    for(int i=1;i<(1<<m);i++)
    {
        v.clear();
        for(int j=0;j<m;j++)
        {
            if(i&(1<<j))
              v.push_back(a[j]);
        }
        lc=1;
        for(int j=0;j<v.size();j++)
            lc=lcm(lc,v[j]);
        if(v.size()&1)
            ans+=n/lc;
        else
            ans-=n/lc;
    }
    cout<<n-ans<<endl;
}
int main()
{
    while(cin>>n>>m)
    {
        REP(i,m)   cin>>a[i];
        solve();
    }
    return 0;
}






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2018-7-20 14:53:51
 
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