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名字是法雷数列其实是欧拉phi函数
Farey Sequence
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 2 Sample Output 1 Source
POJ Contest,Author:Mathematica@ZSU
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#include <iostream>
#include <cstring>
#include <cstdio> using namespace std; typedef unsigned long long int LL; LL sum[],phi[]; void phi_table()
{
phi[]=;
for(int i=;i<=;i++)if(!phi[i])
{
for(int j=i;j<=;j+=i)
{
if(!phi[j]) phi[j]=j;
phi[j]=phi[j]/i*(i-);
}
}
sum[]=;
for(int i=;i<=;i++)
{
sum[i]=sum[i-]+phi[i];
}
} int main()
{
phi_table();
int n;
while(scanf("%d",&n)!=EOF&&n)
{
printf("%llu\n",sum[n]);
}
return ;
}