POJ 2478 Farey Sequence（欧拉函数前n项和）

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A - Farey Sequence

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2478

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 <= 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

Sample Output

题意求欧拉函数的前n项和水题打表筛选即可

#include <iostream>#include <math.h>using namespace std;long long a[1000005]={0};long long c[1000005]={0};void enlur(){    int i,j;    for(i=2;i<1000005;i++)    {        if(!a[i])        {            for(j=i;j<1000005;j=j+i)            {                if(!a[j])                a[j]=j;                a[j]=a[j]/i*(i-1);            }        }    }}int main(){    int i,n;    enlur();    c[0]=a[0];    for(i=1;i<1000005;i++)    c[i]=c[i-1]+a[i];    while(cin>>n&&n)    cout<<c[n]<<endl;    return 0;}

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