The Euler function
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3817 Accepted Submission(s): 1580
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)
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
/*
HODJ 2824 欧拉函数
欧拉函数的定义:E(k)=([1,n-1]中与n互质的整数个数).
(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<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
#define N 3000001
int phi[N];
void phi_table()
{
int i,j;
memset(phi,0,sizeof(phi));
phi[1]=1;
for(i=2;i<=N;i++)
{
if(!phi[i])
{
for(j=i;j<=N;j+=i)
{
if(!phi[j])
phi[j]=j;
phi[j]=phi[j]/i*(i-1);
}
}
}
}
int main()
{
int i,a,b;
__int64 ans;
phi_table();
//freopen("test.txt","r",stdin);
while(scanf("%d%d",&a,&b)!=EOF)
{
ans=0;
for(i=a;i<=b;i++)
ans+=(__int64)phi[i];
printf("%I64d\n",ans);
}
return 0;
}