HDU 2824 The Euler function --------欧拉模板

The Euler function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2986    Accepted Submission(s): 1221

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

 

第一种打表的方法是，素数和欧拉，分开来打表。250ms

第二种打表只有一个，但是时间上更多。500ms

 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;


int prime[3000003],len;
int   opl[3000003];
bool s[3000003];

void Getprime() //打素数表
{
    int i,j;
    len=0;
    for(i=2;i<=3000000;i++)
    {
        if(s[i]==false)
        {
            prime[++len]=i;
            for(j=i*2;j<=3000000;j=j+i)
            s[j]=true;
        }
    }
}

void Euler() //欧拉打表。
{
    int i,j;
    Getprime();
    for(i=2;i<=3000000;i++)
    opl[i]=i;
    opl[1]=0;
    for(i=1;i<=len;i++)
    {
        for(j=prime[i];j<=3000000;j=j+prime[i])
        opl[j]=opl[j]/prime[i]*(prime[i]-1); //利用的定理

    }
}

int main()
{
    int n,m,i;
    __int64 num;
    Euler();
    while(scanf("%d%d",&n,&m)>0)
    {
        num=0;
        for(i=n;i<=m;i++)
        num=num+opl[i];
        printf("%I64d\n",num);
    }
    return 0;
}

 第二种方法。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;


int   opl[3000003];
bool s[3000003];


void Euler() //欧拉打表。
{
    int i,j;
    for(i=2;i<=3000000;i++)
    opl[i]=i;
    opl[1]=0;

    for(i=2;i<=3000000;i++)
    if(s[i]==false)
    {
        for(j=i;j<=3000000;j=j+i)
        {
            opl[j]=opl[j]/i*(i-1);
            s[j]=true;
        }
    }
}

int main()
{
    int n,m,i;
    __int64 num;
    Euler();
    while(scanf("%d%d",&n,&m)>0)
    {
        num=0;
        for(i=n;i<=m;i++)
        num=num+opl[i];
        printf("%I64d\n",num);
    }
    return 0;
}