【SCOI2010】【BZOJ1853】幸运数字

标签：数学   容斥原理   

随便rand出一道题没想到这么可怕QAQ

之前没写过容斥的题

写这个题是第一次

Description

Input

Output

Sample Input

Sample Output

HINT

Source

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
long long L,R;
long long lucky[10001]={0,6,8};
int head=1,tail=2;
long long t[10001];
bool flag[10001];
int num;
long long ans;
long long gcd(long long a,long long b)
{
	if (b) return gcd(b,a%b);
	else return a;
}
void dfs(int now,bool bo,long long x)
{
	if (now>num)
	{
		if (x!=1)
		{	
		    if (bo) ans+=R/x-(L-1)/x;
		    else ans-=R/x-(L-1)/x;
		}
		return;
	}
	dfs(now+1,bo,x);
	long long temp=x/gcd(lucky[now],x);
	double lcm=(double)(temp)*lucky[now];
	if (lcm<=R)
	    dfs(now+1,!bo,temp*lucky[now]);
}
int main()
{
	cin>>L>>R;
	while (lucky[tail]<R)
	{
		lucky[++tail]=lucky[head]*10+6;
		lucky[++tail]=lucky[head]*10+8;
		head++;
	}//先把lucky number搞出来 
	while (lucky[tail]>R) tail--;
	for (int i=1;i<=tail;i++)
	    if (!flag[i])
	    {
	    	t[++num]=lucky[i];
	    	for (int j=i+1;j<=tail;j++)
	    	    if (lucky[j]%lucky[i]==0)
	    	        flag[j]=1;
	    }
	for (int i=1;i<=num;i++) lucky[num-i+1]=t[i];//听说从大到小算可以减少对并集的运算 
	dfs(1,0,1);
	cout<<ans;
}

【SCOI2010】【BZOJ1853】幸运数字

标签：数学   容斥原理   

原文地址：http://blog.csdn.net/creationaugust/article/details/43601517