The Euler function

Link:http://acm.hdu.edu.cn/showproblem.php?pid=2824

The Euler function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3889    Accepted Submission(s): 1613

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

 

大概：求a~b段欧拉值和，要用到筛法欧拉函数。

筛选：1)如果n为素数，φ(n)=n-1;

           2)如果m,n互质，φ(n*m)=φ(n)*φ(m)；

核心就是当m%p==0:

f(m*p) = f(m) * p;//因为（p-1)是在f（m）中所以是乘p；

否则:

f(m*p) = f(m) * (p-1);

因为Euler函数是积性函数所以

综合上面两个式子就可以知道核心的推导公式如何得来

   

 #include<iostream>
 #include<cstdio>
 #include<ctime>
 using namespace std;
 typedef  __int64 INT;
 #define N 3000001
 int prime[216818],phi[N];
 bool a[N];
 void is_prime()
 {
     int i,j,k;
     k=0;
     for(i=2;i<N;i++)
     {
        if(!a[i])
        {
           prime[k++]=i;
           phi[i]=i-1;
        }
        for(j=0;j<k&&i*prime[j]<N;j++)
        {
          a[prime[j]*i]=1;
          if(i%prime[j])
              phi[i*prime[j]]=phi[i]*(prime[j]-1);
          else
          {
              phi[i*prime[j]]=phi[i]*prime[j];
              break;
          }

        }

     }
 }
 int main()
 {
     //freopen("c_in.txt","r",stdin);
     //freopen("myout5.txt","w",stdout);
     int a,b,i;
     INT sum;
     is_prime();
     while(cin>>a>>b)
     {
         sum=0;
         for(i=a;i<=b;i++)
             sum+=phi[i];
         printf("%I64d\n",sum);

     }
     //printf("%lfMS\n",(double)clock()/CLOCKS_PER_SEC);
     return 0;
 }