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

题目地址：点击打开链接

题意：给定n和一个大小为m的集合，集合元素为非负整数。为1…n内能被集合里任意

一个数整除的数字个数。n<=2^31,m<=10

解题思路：容斥原理地简单应用。先找出1...n内能被集合中任意一个元素整除的个数，

再减去能被集合中任意两个整除的个数，即能被它们两只的最小公倍数整除的个数，因为

这部分被计算了两次，然后又加上三个时候的个数，然后又减去四个时候的倍数...所以深

搜，最后判断下集合元素的个数为奇还是偶，奇加偶减。

注意：可能有0，要先去掉

代码：

#include<bits/stdc++.h>using namespace std;typedef long long ll;ll n, m, a[25], cnt, ans, t;void dfs(int cur, ll lcm, int id){    lcm = a[cur]/__gcd(lcm, a[cur])*lcm;    if(id%2) ans += (n-1)/lcm;    else ans -= (n-1)/lcm;    for(int i = cur+1; i < cnt; i++)        dfs(i, lcm, id+1);}int main(void){    while(cin >> n >> m)    {        ans = 0, cnt = 0;        for(int i  = 0; i < m; i++)        {            scanf("%lld", &t);            if(t) a[cnt++] = t;        }        for(int i = 0; i < cnt; i++)            dfs(i, a[i], 1);        printf("%lld\n", ans);    }    return 0;}

 

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7554    Accepted Submission(s): 2233

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 22 3

 

Sample Output

7

 

Author

wangye

 

Source

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

 

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