POJ2478 Farey Sequence(欧拉函数,打表)

Farey Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13435 Accepted: 5272

Description

The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few are 
F2 = {1/2} 
F3 = {1/3, 1/2, 2/3} 
F4 = {1/4, 1/3, 1/2, 2/3, 3/4} 
F5 = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5} 

You task is to calculate the number of terms in the Farey sequence Fn.

Input

There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 10⁶). There are no blank lines between cases. A line with a single 0 terminates the input.

Output

For each test case, you should output one line, which contains N(n) ---- the number of terms in the Farey sequence Fn. 

Sample Input

23450

Sample Output

1359

题目大意:

法雷级数Fn(n >= 2)由一系列不能约分的分数a/b(0 < a < b  <= n 且gcd(a,b) =1)按递增的顺序排列组成,下面是法雷级数的前几项:

F2 = {1/2}

F3 = {1/3,1/2,2/3}

F4 = {1/4,1/3,1/2,2/3,3/4}

F5 = {1/5,1/4,1/3,2/5,1/2,3/5,2/3,3/4,4/5}

解题思路:

一个法雷序列Fn中数的个数就是分别与2,3,4,5,....,n-1,n互素的数的个数和,继而就是求从2到n连续的欧拉函数值的和,因为N的范围是[2,1000000],可以直接用欧拉函数的递推方法来求解，再打表预处理前n项和即可。

AC代码：

#include<iostream>#include<cstdio>using namespace std;const int maxn = 1000001;long long phi[maxn];void dabiao(){int i,j;for(i=1;i<=maxn;i++)phi[i] = i;for(i=2;i<=maxn;i+=2)phi[i] /= 2;for(i=3;i<=maxn;i+=2)if(phi[i] == i){for(j=i;j<=maxn;j+=i){phi[j] = phi[j] / i * (i-1);}}for(i=3;i<=1000001;i++){phi[i] += phi[i-1]; }}int main(){int m;dabiao();while(scanf("%d",&m) != EOF && m){printf("%lld\n",phi[m]);}return 0;｝