HDU 2824 The Euler function（欧拉函数打表）

最新推荐文章于 2019-09-03 21:18:17 发布

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最新推荐文章于 2019-09-03 21:18:17 发布

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本文深入探讨欧拉函数的概念，结合实例解析其性质，并提供一种打表方法用于计算欧拉函数值，帮助理解该数论函数在数论问题中的应用。

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The Euler function

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8579 Accepted Submission(s): 3598

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

#include <stdio.h>  
#define Max 3000001  
typedef long long ll;  
using namespace std;  
ll euler[Max];  
void Init()
{   
    euler[1]=1;  
    for(int i=2;i<Max;i++)  
        euler[i]=i;  
    for(int i=2;i<Max;i++)  
       if(euler[i]==i)  
            for(int j=i;j<Max;j+=i)  
            	euler[j]=euler[j]/i*(i-1);  
    //先进行除法是为了防止中间数据的溢出,以上就是打表  
    for(int i=2;i<Max;i++)
		euler[i]+=euler[i-1];  
    //将[1,i-1]的和叠加到euler[i]上可以节省一个数组，这句根据题意的改变  
}  
int main()  
{  
    Init();  
    int a,b;  
    while(~scanf("%d %d",&a,&b))
	{
		printf("%I64d\n",euler[b]-euler[a-1]);
	}   
    return 0;  
}