杭电2824--The Euler function（欧拉函数）

The Euler function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4559    Accepted Submission(s): 1902

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

3042

 

 

Source

2009 Multi-University Training Contest 1 - Host by TJU

 

 

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欧拉函数是指：对于一个正整数n，小于n且和n互质的正整数（包括1）的个数，记作φ(n) 。

通式：φ(x)=x*(1-1/p1)*(1-1/p2)*(1-1/p3)*(1-1/p4)…..(1-1/pn),其中p1, p2……pn为x的所有质因数，x是不为0的整数。φ(1)=1（唯一和1互质的数就是1本身）。

对于质数p，φ(p) = p - 1。注意φ(1)=1.

欧拉定理：对于互质的正整数a和n，有aφ(n) ≡ 1 mod n。

欧拉函数是积性函数——若m,n互质，φ(mn)=φ(m)φ(n)。

                                 若n是质数p的k次幂，φ(n)=p^k-p^(k-1)=(p-1)p^(k-1)，因为除了p的倍数外，其他数都跟n互质。

特殊性质：当n为奇数时，φ(2n)=φ(n)

欧拉函数还有这样的性质：

设a为N的质因数，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 <cstdio>
#include <cstring>
#include <iostream>
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++){
            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()
{
    int a, b;
    Get_phi();
    while(~scanf("%d %d", &a, &b))
    {
        long long sum = 0;
        for(int i = a; i <= b; i++)
            sum += phi[i];
        cout << sum << endl;
    }
    return 0;
}