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

最新推荐文章于 2024-08-17 03:37:44 发布

原创最新推荐文章于 2024-08-17 03:37:44 发布 · 241 阅读

1 ·

CC 4.0 BY-SA版权

数论专栏收录该内容

9 篇文章

订阅专栏

本文解析了一道经典的算法题目：给定一个数N和一个整数集合，找出小于N的所有整数中能被集合中任意整数整除的数量。通过去重、求最小公倍数并使用容斥原理解决此问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

How many integers can you find
Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10614 Accepted Submission(s): 3159

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）

Recommend

wangye

【思路】

先去重，然后求不同数的组合（利用二进制思想）得到的最小公倍数，容斥一下即可，注意要求的是比N小的那些数的个数，所以N能整除的要去掉。

【代码】

//******************************************************************************
// File Name: HDU_1796.cpp
// Author: Shili_Xu
// E-Mail: shili_xu@qq.com
// Created Time: 2018年08月05日 星期日 10时21分43秒
//******************************************************************************

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;

set<int> st;
int a[25];
int n, m;

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

ll lcm(ll a, ll b)
{
	return a * b / gcd(a, b);
}

int main()
{
	while (scanf("%d %d", &n, &m) == 2) {
		int x;
		st.clear();
		for (int i = 1; i <= m; i++) scanf("%d", &x), st.insert(x);
		m = 0;
		for (set<int>::iterator it = st.begin(); it != st.end(); ++it)
			if (*it) a[++m] = *it;
		ll ans = 0, all = 1 << m;
		for (int i = 1; i < all; i++) {
			int cnt = 0, now = 1;
			for (int j = 0; j < m; j++)
				if (i & (1 << j)) cnt++, now = lcm(now, a[j + 1]);
			ans += (cnt & 1 ? 1 : -1) * (n / now);
			if (n % now == 0) ans -= (cnt & 1 ? 1 : -1);
		}
		printf("%lld\n", ans);
	}
	return 0;
}