HDU 1796 How many integers can you find（简单容斥原理）

题目大意：有一个序列，大小为m，里面有m个不超过20的非负数，各不相同。要求在1-n中有多少个能被m个数中任意一个数整除。

题目思路：简单的容斥原理应用。就不说了直接上代码。

有两种方法，一种是DFS，一种是直接位元素枚举暴力（study from zhixiaoli)

DFS：（速度较快）

#include<iostream>
#include<algorithm>
using namespace std;
long long a[15];
int n, m;
int gcd(int a, int b)
{
	if (b == 0)
		return 0;
	int c = a%b;
	while (c)
	{
		a = b;
		b = c;
		c = a%b;
	}
	return b;
}
long long lcm(long long a, long long b)
{
	long long c = gcd(a, b);
	if (c == 0)
		return 0;
	return a*b / c;
}
int gao(int x)
{
	if (x == 0)
		return 0;
	return n / x;
}
int dfs(int i, bool flag, long long LCM)
{
	long long ans = 0;
	long long mid = lcm(max(LCM,a[i]), min(LCM, a[i]));
	if (flag) ans += gao(mid);
	else ans -= gao(mid);
	for (int j = i + 1; j <= m; j++)
		ans += dfs(j, !flag, mid);
	return ans;
}
int main()
{
	while (cin >> n >> m)
	{
		n--;
		for (int i = 1; i <= m; i++)
			cin >> a[i];
		long long num = 0;
		for (int i = 1; i <= m; i++)
			num += dfs(i, 1, 1);
		cout << num << endl;
	}
}

位元素枚举：（速度较慢）

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
using namespace std;
#define ll long long
int n, m;
ll a[15];
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;
}
ll gao(int l, int&cnt)
{
	ll res = 1;
	for (int i = cnt = 0; i < m; i++)
	{
		if (l&(1 << i))
		{
			cnt++;
			res = LCM(res, a[i]);
		}
	}
	return res;
}
int main()
{
	while (~scanf("%d%d", &n, &m))
	{
		n--;
		int k=m;
		for (int i = 0; i < m; i++)
		{
			scanf("%lld", &a[i]);
			if (!a[i]) k = i;
		}
		if (k != m)
		{
			m--;
			swap(a[m], a[k]);
		}
		ll ans = 0;
		for (int i = 1; i < (1 << m); i++)
		{
			ll temp = gao(i, k);
			if (k & 1)
				ans += n / temp;
			else
				ans -= n / temp;
		}
		cout << ans << endl;
	}
}