hdoj2824 The Euler function（欧拉函数）

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

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
欧拉函数介绍：
先介绍一下暴力的欧拉函数：
Eular(m) = m - (1-1/p1) - (1-1/p2) - ... - (1-1/pk)  [其中 p1, p2...pk为m的素因子]
1.  phi(p) == p-1 因为素数p除了1以外的因子只有p，所以与 p 互素的个数是 p - 1个

2. phi(p^k) == p^k - p^(k-1) == (p-1) * p^(k-1)
证明：
令n == p^k,小于 n 的正整数共有 p^k-1 个,其中与 p 不互素的个数共 p^(k-1)-1 个，它们是 1*p,2*p,3*p ... (p^(k-1)-1)*p
所以phi(p^k) == (p^k-1) - (p^(k-1)-1) == p^k - p^(k-1) == (p-1) * p^(k-1)。

3. 如果i mod p == 0, 那么 phi(i * p) == p * phi(i) （证明略）
举个例子：
假设 p = 3,i = 6,p * i = 18 = 2 * 3^2;
phi(3 * 6) == 18*(1-1/2)*(1-1/3) = 6
p * phi(i) = 3 * phi(6) = 3 * 6 * (1-1/2) *  (1-1/3) = 6 = phi(i * p) 正确

4. 如果i mod p != 0, 那么 phi(i * p) == phi(i) * (p-1) 
证明：
i mod p 不为0且p为质数, 所以i与p互质, 那么根据积性函数的性质 phi(i * p) == phi(i) * phi(p) 其中phi(p) == p-1
所以 phi(i * p) == phi(i) * (p-1).
再举个例子：
假设i = 4, p = 3, i * p = 3 * 4 = 12
phi(12) = 12 * (1-1/2) * (1-1/3) = 4
phi(i) * (p-1) = phi(4) * (3-1) = 4 * (1-1/2) * 2 = 4 = phi(i * p)正确
打表有两种方法，一种是直接从公式入手，另一种是从欧拉函数的定义入手。
方法一（从公式入手）
代码如下：
#include<cstdio>#include<algorithm> #define  MAXN 3000300int euler[MAXN]; void init(){    euler[1]=1;    for(int i=1;i<MAXN;i++)        euler[i]=i;    for(int i=2;i<MAXN;i++)    {        if(euler[i]==i)        {        //printf("%d\n",i);            for(int j=i;j<MAXN;j+=i)                euler[j]=euler[j]*(i-1)/i;        }    }}int main(){init();int a,b;while(scanf("%d%d",&a,&b)!=EOF){__int64 ans=0;for(int i=a;i<=b;i++)ans+=euler[i];printf("%I64d\n",ans);}return 0;}
方法二：（从性质定义入手）

代码如下：
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define   maxn  3000300int flag[maxn];int phi[maxn];int p[maxn];int get_phi(){int i,j;memset(flag,1,sizeof(flag));int l=0;phi[1]=1; for(i=2;i<maxn;i++){if(flag[i]){p[l++]=i;phi[i]=i-1;}for(j=0;j<l;j++){if(i*p[j]>maxn)break;flag[i*p[j]]=0;if(i%p[j]==0){  phi[i*p[j]]=phi[i]*p[j];  break;}  else  phi[i*p[j]]=phi[i]*(p[j]-1);} } } int main(){get_phi();/*for(int i=2;i<=100;i++)printf("%d\n",phi[i]);*/__int64 a,b;while(scanf("%d%d",&a,&b)!=EOF){__int64 ans=0;for(int i=a;i<=b;i++){ans+=phi[i];}printf("%I64d\n",ans);}return 0;}