hdu2824（欧拉函数）

The Euler function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3674    Accepted Submission(s): 1509

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

 

总结：最初，我用sum[i]数组存储1到i的欧拉函数值得和，但是因为开了两个数组，超了空间，本想用此法优化时间的，结果还是要从a循环到b取每个数的欧拉函数的值相加

#include<stdio.h>
#include<string.h>
__int64 euler[3000000];
int main()
{
    __int64 ans;
    memset(euler,0,sizeof(euler));
    euler[1] = 1;
    int a,b,i,j;
    for(i = 2; i <3000000; i++)
    {
        if(!euler[i])
            for(j = i; j <3000000; j += i)
            {
                if(!euler[j])
                    euler[j] = j;
                euler[j] = euler[j]/i*(i-1);
            }
    }
    while(scanf("%d%d",&a,&b)!=EOF)
    {
        ans=0;
        for(i=a; i<=b; i++)
            ans+=euler[i];
        printf("%I64d
",ans);
    }
    return 0;
}