hdu 1796 容斥原理（dfs）

很基本的容斥原理的题目，注意要先把0去掉。

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

typedef long long ll;
const int N = 10;
int s[N];
bool visit[N];
int n, m, ans;

ll gcd( ll x, ll y )
{
    return y ? gcd( y, x % y ) : x;
}

ll lcm( ll x, ll y )
{
    return x / gcd( x, y ) * y;
}

void dfs( int cur )
{
    if ( cur == m )
    {
        int cnt = 0, r = 1;
        ll d = 1;
        for ( int i = 0; i < m; i++ )
        {
            if ( visit[i] )
            {
                d = lcm( d, s[i] );
                cnt++;
            }
        }
        if ( cnt == 0 ) return ;
        if ( cnt % 2 == 0 ) r = -1;            
        ans += n / d * r;
        return ;
    }    
    visit[cur] = 0;
    dfs( cur + 1 );
    visit[cur] = 1;
    dfs( cur + 1 );
    return ;
}

int main ()
{
    while ( scanf("%d%d", &n, &m) != EOF )
    {
        int tn = 0;
        for ( int i = 0; i < m; i++ )
        {
            int tmp;
            scanf("%d", &tmp);
            if ( tmp != 0 )
            {
                s[tn++] = tmp;
            }
        }
        m = tn;
        ans = 0;
        n--;
        dfs(0);
        printf("%d
", ans);
    }
    return 0;
}