欧拉函数

打巧克力杯周赛的时候，遇到了一道欧拉函数板子题，然而我居然不会（数论全推给队友），结果是我居然用埃氏筛自己打出来了，居然还是对的。

下面是那道题：

例题：

Farey Sequence Length

Given a positive integer, NN, the sequence of all fractions a/ba/b with 0≤a≤b0≤a≤b, 1≤b≤N1≤b≤N and aa and bb relatively prime, listed in increasing order, is called the Farey Sequence of order NN. For example, the Farey Sequence of order 66 is:

0/1,1/6,1/5,1/4,1/3,2/5,1/2,3/5,2/3,3/4,4/5,5/6,1/10/1,1/6,1/5,1/4,1/3,2/5,1/2,3/5,2/3,3/4,4/5,5/6,1/1

For this problem, you will write a program to compute the length of the Farey Sequence of order NN.

Input

The first line of input contains a single integer PP (1≤P≤100001≤P≤10000), which is the number of data sets that follow. Each data set should be processed identically and independently.

Each data set consists of a single line of input. It contains the data set number, KK, followed by the order NN (2≤N≤100002≤N≤10000) of the Farey Sequence whose length is to be found.

Output

For each data set there is a single line of output. The single output line consists of the data set number, KK, followed by a single space followed by the length of the Farey Sequence as a decimal integer.

Sample Input 1	Sample Output 1
4 1 6 2 15 3 57 4 9999	1 13 2 73 3 1001 4 30393487

 

题意：就是让你求欧拉函数前n项和。

其实正解也是用埃氏筛做的。

比赛代码：

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

int a[10005];
int ans[10005];

int main()
{
	ans[0]=0;
	for (int i=1;i<=10000;i++)
	{
		a[i]=a[i]+i+1;
		for (int j=i*2;j<=10000;j+=i)
			a[j]-=a[i];
		ans[i]=ans[i-1]+a[i];
	}
	int P,K,N;
	cin >> P;
	while(P--)
	{
		scanf("%d%d",&K,&N);
		printf("%d %d\n",K,ans[N]);
	}
	return 0;
}

 

直接贴模板了，具体公式百度。

欧拉函数模板：

求单个欧拉函数：

int phi(int n)
{
	int res=n;
	for(int i=2;i*i<=n;i++)
		if(n%i==0)
		{
			res=res-res/i;
			while(n%i==0) n/=i;
		}
	if(n>1)
		res=res-res/n;
	return res;
}

埃氏筛：

const int SIZE=10000+5;
int phi[SIZE];

void init()
{
	memset(phi,0,sizeof(phi));
	phi[1]=1;
	for (int i=2;i<SIZE;i++)
	if(!phi[i])
	{
		for (int j=i;j<SIZE;j+=i)
		{
			if(!phi[j]) phi[j]=j;
			phi[j]=phi[j]/i*(i-1);
		}
	}
}