uestc 1720 square-free integer(无平方因子数)&nyoj 580square-free integer（容斥，数论）

无平方因子数即对于任意一个素数p，p^2都不会整除那个数，如，3,5,7,15=3*5都是无平方因子数，而20=4*5=2^2*5不是！

先找素数，然后用容斥原理，递归求解

描述

In mathematics, a square-free integer is one divisible by no perfect square, expect 1.Such as 10,but 12 is not because 12=4×3.

So the problem is very simple, give you an interval [n,m],can should calculate how many square-free numbers in this interval?

输入

Each line will consist of two integers n,m(0<n<=m<10^9),end by EOF.

输出

Out put each answer in one line.

样例输入

1 10
10 20

样例输出

7
7

java代码nyoj 580：

 
import java.util.Scanner;
public class Main {
	static int size=(int)Math.sqrt(999999999);
	static int prime[]=new int[size+10];
	static boolean flag[]=new boolean[size+10];
	static int count=0;
	public static void prim()
	{
		for(int i=2;i<=size;i++)
		{
			if(!flag[i])
			{
				prime[count++]=i;
				for(int j=2*i;j<=size;j+=i)
				{
					flag[j]=true;
				}
			}
		}
	}
	
	//容斥原理
	public static int dfs(int index,int num)
	{
		int sum=0;
		for(int i=index;i<count&&(prime[i]*prime[i])<=num;i++)
		{
			sum+=num/prime[i]/prime[i]-dfs(i+1, num/prime[i]/prime[i]);
		}
		return sum;
	}
	public static void main(String[] args) {
		prim();
		Scanner scanner=new Scanner(System.in);
		while(scanner.hasNext())
		{
			int n=scanner.nextInt();
			int m=scanner.nextInt();
			System.out.println(((m-dfs(0, m))-(n-dfs(0, n)))+1);
		}

	}

}
        

uestc  ac代码：

#include<stdio.h>
#define  size 1000005
int prime[size];
bool flag[size];
int count=0;
	 void prim()
	{
		for(int i=2;i<=size;i++)
		{
			if(!flag[i])
			{
				prime[count++]=i;
				for(int j=2;j*i<size;j++)
				{
					flag[j*i]=1;
				}
			}
		}
	}
	
	//容斥原理
	long long int dfs(int index,long long int num)
	{
		long long int sum=0;
		for(int i=index;i<count&&(long long int)prime[i]*prime[i]<=num;i++)
		{
			sum+=num/prime[i]/prime[i]-dfs(i+1, num/prime[i]/prime[i]);
		}
		return sum;
	}
int main()
{
	int cases;
	prim();
	scanf("%d",&cases);
	while(cases--)
	{
        long long int n;
		scanf("%lld",&n);
		printf("%lld\n",n-dfs(0,n));
	}
	return 0;
}