HDU-1796 How many integers can you find (容斥定理+dfs)

How many integers can you find
Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13231 Accepted Submission(s): 3968

Problem Description
Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10}, all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7.

Input
There are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.

Output
For each case, output the number.

Sample Input
12 2
2 3

Sample Output
7

Author
wangye

Source
2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）

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题意：

给出两个数N(0<N<2³¹)和M(0<M<=10)，M是一个不超过10个元素的非负整数集合，元素值不超过20. 求出在小于N的整数中满足“能整除M集合里其中一个元素”的元素个数。
比如N=12, M{2,3}, 满足的集合为{2,3,4,6,8,9,10}，该集合元素个数为7，就输出7.

思路分析：

分析样例，11里面整除2的为2,4,6,8,10. 个数为[11/2]。11里面整除3的为3,6,9. 个数为[11/3]。
其中，6被重复计算了一次，需要减掉。这个6就是2和3的最小公倍数。因此需要减掉[11/6]。
可以发现这种计算模式就是容斥定理。相交部分就是最小公倍数。

容斥定理核心概念为“奇加偶减”。

举个例子，N=15, M = 3, 集合为{2,5,7}，[a,b]表示lcm(a,b)，那么ans = f(2) + f(5) + f(7) - f([2,5]) - f([2,7]) - f([5,7]) + f([2,5,7]).
我们改变一下排列方式
2
2 5
2 5 7
2 7
5
5 7
7

这时可以很明显的发现这个就是类似于子集生成的dfs，个数为奇数时加，个数为偶数时减。

下面上AC代码

#include <iostream>
#define ll long long 
using namespace std;

ll ans, N, M, a[100];

ll gcd(ll a,ll b){
    return b == 0?a:gcd(b,a%b);
}

void dfs(ll cur, ll lcm, ll id)
{
	lcm = a[cur]*lcm/gcd(lcm,a[cur]);
//	cout<<cur<<" "<<lcm<<endl;
	if(id&1) ans += (N-1)/lcm;	//读取id最后一个二进制位，若为1，则id为奇数。 
	else ans -= (N-1)/lcm;
	for(int i = cur+1;i < M;i++)
	dfs(i, lcm, id+1);
}

int main()
{
	while(cin>>N>>M)
	{
		ans = 0;
		ll t = 0;
		ll MM = M;
		while(MM--)
		{
			ll x;
			cin>>x;
			if(x) a[t++] = x;	//去除0 
		}
		for(int i = 0;i < M;i++)
		dfs(i,a[i],1);
		cout<<ans<<endl;
	}
	return 0;
}