A - 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}
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
题意：求欧拉函数的前n项和。
题解：打表筛选即可。
#include <iostream>
using namespace std;
typedef long long ll;
const int maxn=1e6+;
ll a[maxn];
void get()
{
for(int i=;i<maxn;i++)
{
if(!a[i])
{
for(int j=i;j<maxn;j+=i) //已经包含了素筛的思想
{
if(!a[j])
a[j]=j;
a[j]=a[j]/i*(i-);
}
}
a[i]+=a[i-];
}
}
int main()
{
get();
int n;
while(cin>>n&&n)
cout<<a[n]<<endl;
return ;
}