UVA 10325 The Lottery(容斥），10325lottery

UVA 10325 The Lottery(容斥），10325lottery

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 ProblemThere 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 22 320 22 4

Sample Output

310

Md. Kamruzzaman把出局的人找出來拿n減去出局的人就可以了。考慮容斥做法。設f[i]表示i個數的最小公倍數。那麼出局總人數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;}