Hduoj2824【欧拉函数】

/*The Euler function
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4299    Accepted Submission(s): 1788


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
 
Source
2009 Multi-University Training Contest 1 - Host by TJU 

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*/ 
#include<stdio.h>
__int64 e[3000010];
int main()
{
	int i, j, k, n, m;
	for(i = 1; i <= 3000005; ++i)
	e[i] = i;
	for(i = 2; i <= 3000005; ++i)
	{
		if(e[i] == i)
		for(j = i; j <= 3000005; j += i)
		e[j] = e[j]/i*(i-1);
	}
	for(i = 2; i <= 3000005; ++i)
	e[i] += e[i-1];
	while(scanf("%d%d", &n, &m) != EOF)
	{
		printf("%I64d\n", e[m] - e[n-1]);
	}
	return 0;
} 

题意：给出a和b求出a到b中所有数的欧拉函数的和。

思路：就是最基本的欧拉函数模板题，要根据欧拉函数的定义e【i】=i * (1 - 1 / p1) * （1 - 1 / p2) ......* (1 - 1 / pn) ，其中pn为i的质因数。