【HDU】2824 - GCD（欧拉函数打表）

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The Euler function

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5451 Accepted Submission(s): 2307

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

打一个 1 到 MAX 的欧拉函数表，表示 1 加到 i 的欧拉函数值。

代码如下：

#include <cstdio>
#define MAX 3000000
__int64 eu[MAX+5];
void Eular()
{
	eu[1] = 1;
	for (int i = 2 ; i <= MAX ; i++)
	{
		if (!eu[i])
		{
			for (int j = i ; j <= MAX ; j += i)
			{
				if (!eu[j])
					eu[j] = j;
				eu[j] -= eu[j] / i;
			}
		}
		eu[i] += eu[i-1];
	}
}
int main()
{
	Eular();
	int a,b;
	while (~scanf ("%d %d",&a,&b))
		printf ("%I64d\n",eu[b] - eu[a-1]);
	return 0;
}