poj 3090 && poj 2478（法雷级数，欧拉函数）

http://poj.org/problem?id=3090

法雷级数

法雷级数的递推公式非常easy：f[1] = 2; f[i] = f[i-1]+phi[i]。

该题是法雷级数的变形吧，答案是2*f[i]-1。

#include <stdio.h>
#include <iostream>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL long long
#define _LL __int64
#define eps 1e-12
#define PI acos(-1.0)
using namespace std;

const int maxn = 1100;

int flag[maxn];
int prime[maxn];
int phi[maxn];
LL f[maxn];

void init()
{
	memset(flag,0,sizeof(flag));
	prime[0] = 0;
	phi[1] = 1;
	for(int i = 2; i < maxn; i++)
	{
		if(flag[i] == 0)
		{
			phi[i] = i-1;
			prime[++prime[0]] = i;
		}
		for(int j = 1; j <= prime[0]&&prime[j]*i<maxn; j++)
		{
			flag[prime[j]*i] = 1;
			if(i % prime[j] == 0)
				phi[prime[j]*i] = phi[i] * prime[j];
			else
				phi[prime[j]*i] = phi[i] * (prime[j] - 1);
		}
	}
	f[1] = 2;
	for(int i = 2; i <= 1000; i++)
		f[i] = f[i-1] + phi[i];
}

int main()
{
	init();
	int test;
	scanf("%d",&test);
	for(int item = 1; item <= test; item++)
	{
		int x;
		scanf("%d",&x);
		printf("%d %d %lld\n",item,x,f[x]*2-1);
	}
	return 0;
}

id=2478">http://poj.org/problem?

id=2478

更简单了，直接求法雷级数。基于素数筛的欧拉函数。

#include <stdio.h>
#include <iostream>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL long long
#define _LL __int64
#define eps 1e-12
#define PI acos(-1.0)
using namespace std;

const int maxn = 1000010;

int flag[maxn];
int prime[maxn];
int phi[maxn];
LL f[maxn];

void init()
{
	memset(flag,0,sizeof(flag));
	prime[0] = 0;
	phi[1] = 1;
	for(int i = 2; i < maxn; i++)
	{
		if(flag[i] == 0)
		{
			phi[i] = i-1;
			prime[++prime[0]] = i;
		}
		for(int j = 1; j <= prime[0]&&prime[j]*i<maxn; j++)
		{
			flag[prime[j]*i] = 1;
			if(i % prime[j] == 0)
				phi[prime[j]*i] = phi[i] * prime[j];
			else
				phi[prime[j]*i] = phi[i] * (prime[j] - 1);
		}
	}
	f[1] = 2;
	for(int i = 2; i <= 1000010; i++)
		f[i] = f[i-1] + phi[i];
}

int main()
{
	init();
	int n;
	while(~scanf("%d",&n)&&n)
	{
		printf("%lld\n",f[n]-2);
	}

	return 0;
}