HDU 2824 The Euler function

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

题解:

求一个欧拉函数的和,求法参见这篇博客.

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <cstdio>
 6 #include <cmath>
 7 using namespace std;
 8 typedef long long ll;
 9 const int N=3000005,M=300005;
10 bool d[N];int l,r,phi[N],prime[M],m=0;
11 void work(){
12     int tmp;
13     phi[1]=1;
14     for(int i=2;i<N;i++){
15         if(!d[i]){
16             phi[i]=i-1;
17             prime[++m]=i;
18         }
19         for(int j=1;j<=m && i*prime[j]<N;j++){
20             tmp=i*prime[j];
21             d[tmp]=true;
22             if(i%prime[j]){
23                 phi[tmp]=phi[i]*(prime[j]-1);
24             }
25             else{
26                 phi[tmp]=phi[i]*prime[j];
27                 break;
28             }
29         }
30     }
31 }
32 int main()
33 {
34     
35     work();
36     int a,b;
37     while(~scanf("%d%d",&a,&b)){
38         long long sum=0;
39         for(int i=a;i<=b;i++)
40             sum+=phi[i];
41         printf("%lld
",sum);
42     }
43     return 0;
44 }