poj 2478 Farey Sequence

B - Farey Sequence

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

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

2
3
5
4
0

Sample Output

1
3
9
5

这道题的题意可以转化为对每一个数字输出它的上一位sum值加上不超过它的互素整数的和。

求每一位数的不超过它自身的互质整数用欧拉函数：

欧拉函数模板：

int a[n];

void  euler(int n){

memset(a,0,sizeof(a));

a[i]=1;

for(int i=2;i<=n;i++)

      if(!a[i])

for(int j=i;j<=n;j+=i)

{ 

if(!a[j]) a[j]=j;

a[j]=a[j]/i*(i-1);

}

}

注意用 long long 开始用int sum[Max] WA了好几次。

代码：

#include<iostream>#include<cstring>#include<string>using namespace std;int a[1000005];long long sum[1000005];int n;void phi_table(int n){    for(int i=2;i<=n;i++)    a[i]=0;    a[1]=1;    for(int i=2;i<=n;i++)    {if(!a[i])    for(int j=i;j<=n;j+=i)    {        if(!a[j]) a[j]=j;        a[j]=a[j]/i*(i-1);    }    }}int main(){    phi_table(1000000);while(cin>>n&&n!=0)    {        memset(sum,0,sizeof(sum));        for(int i=2;i<=n;i++)        sum[n]=sum[n]+a[i];        cout<<sum[n]<<endl;    }    return 0;}