The Euler function（hdoj --2824-欧拉函数）

The Euler function

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4518 Accepted Submission(s): 1879

Problem Description

The Euler function phi is an important kind of function in number theory, (n) represents the amount of the numbers which are smaller than n and coprime to n, and this function has a lot of beautiful characteristics. Here comes a very easy question: suppose you are given a, b, try to calculate (a)+ (a+1)+....+ (b)

Input

There are several test cases. Each line has two integers a, b (2<a<b<3000000).

Output

Output the result of (a)+ (a+1)+....+ (b)

Sample Input

3 100

Sample Output

Source

2009 Multi-University Training Contest 1 - Host by TJU

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解析：（转）

定义：对于正整数n，φ(n)是小于或等于n的正整数中，与n互质的数的数目。

　　　　例如：φ(8)=4，因为1，3，5，7均和8互质。

性质：1.若p是质数，φ(p)= p-1.

　　　2.若n是质数p的k次幂，φ(n)=(p-1)*p^(k-1)。因为除了p的倍数都与n互质

　　　3.欧拉函数是积性函数，若m,n互质，φ(mn)= φ(m)φ(n).

　　根据这3条性质我们就可以推出一个整数的欧拉函数的公式。因为一个数总可以写成一些质数的乘积的形式。

　　E(k)=(p1-1)(p2-1)...(pi-1)*(p1^(a1-1))(p2^(a2-1))...(pi^(ai-1))

　　　　= k*(p1-1)(p2-1)...(pi-1)/(p1*p2*...*pi)

　　　　= k*(1-1/p1)*(1-1/p2)...(1-1/pk)

在程序中利用欧拉函数如下性质，可以快速求出欧拉函数的值（a为N的质因素）

　　若( N%a ==0&&(N/a)%a ==0)则有：E(N)= E(N/a)*a;

　　若( N%a ==0&&(N/a)%a !=0)则有：E(N)= E(N/a)*(a-1);

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=3000010;
int prime[N],isprime[N];
int phi[N];
void get_phi()
{
    int i,j,cnt=0;
    for(i=2;i<N;i++)
	{
        if(isprime[i]==0)
		{
            prime[cnt++]=i;
            phi[i]=i-1;
        }
        for(j=0;j<cnt && i*prime[j]<N;j++)
		{     //注意这里，i*prime[j]<N 可换成 prime[j]<=N/i(带等号)
            isprime[i*prime[j]]=1;
            if(i%prime[j]==0)
                phi[i*prime[j]]=phi[i]*prime[j];
            else
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
        }
    }
}
int main()
{
    long long sum;
    int a,b;
    get_phi();
    while(~scanf("%d%d",&a,&b))
	{
        sum=0;
        for(int i=a;i<=b;i++)
        sum+=phi[i];
        cout<<sum<<endl;
    }
    return 0;
}