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    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 <= 10 6). 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

    2
    3
    4
    5
    0

    Sample Output

    1
    3
    5
    9

    欧拉函数打表模板

    复杂度O(n*loglogn)

    #include<stdio.h>
    using namespace std;
    long long phi[1000005];
    void getphi(long long maxn){
        for(int i = 2; i < maxn; i++){
            phi[i] = i;
        }
        phi[1] = 1;
        for(int i = 2; i < maxn; i++){
            if(phi[i] == i){
                for(long long j = i; j < maxn; j += i){
                    phi[j] = phi[j] / i * (i - 1);
                }
            }
        }
    }
    long long sum[1000005];
    int main()
    {
        sum[0]=0;
        getphi(1000003);
        for(long long i=1;i<=1000000;i++)
        {
            sum[i]=sum[i-1]+phi[i];
        }
        int n;
        while(~scanf("%lld",&n))
        {
            if(n==0) break;
            printf("%lld
    ",sum[n]-1);
        }
    }
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  • 原文地址:https://www.cnblogs.com/caowenbo/p/11852234.html
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