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你好,容斥原理---层叠消融

程序员文章站 2024-02-11 13:23:28
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http://acm.hdu.edu.cn/showproblem.php?pid=1796 

  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.

  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

你好,容斥原理---层叠消融 

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<cmath>
#include<string>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
//层叠消融---偶消奇不消 
ll num[11]; 
ll m,n;
ll sum=0,cnt;
ll gcd(ll a,ll b){
	return b==0?a:gcd(b,a%b);
}
ll LCM(ll a,ll b){
	return a/gcd(a,b)*b;
}

void dfs(int i,ll lcm,ll step){
	if(step&1)//如果是层叠奇数次 
		sum+=(n-1)/lcm;
	else sum-=(n-1)/lcm;//否则消去 
	for(int k=i+1;k<cnt;k++){
		dfs(k,LCM(lcm,num[k]),step+1);
	}
}

int main(){
	ll x;
	while(cin>>n>>m){
	cnt=1;
	for(int i=1;i<=m;i++){
		cin>>x;
		if(x>0&&x<n)num[cnt++]=x;
	}
	sum=0;
	for(int i=1;i<cnt;i++){
		dfs(i,num[i],1);
	}
	cout<<sum<<endl;
	}
return 0;
}

 

 

 

 

 

相关标签: 容斥原理