poj 3090 && poj 2478（法雷级数，欧拉函数）

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

法雷级数

法雷级数的递推公式很简单：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];elsephi[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;}

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];elsephi[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;}

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