题目链接:http://poj.org/problem?id=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 <= 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
3
4
5
0
Sample Output
1
3
5
9
题目大意:一个数列:
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}
......
观察这个数列很容易就发现规律,第n个数列的元素个数是第n-1数列的元素个数+n的欧拉函数值
给出一个数n,输出该数列的元素个数
很简单,打个表,暴力预处理一下,直接输出
ac:
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<map>
//#include<set>
#include<deque>
#include<queue>
#include<stack>
#include<bitset>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
#define mod 1000000007
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b)
#define clean(a,b) memset(a,b,sizeof(a))// 水印
//std::ios::sync_with_stdio(false);
ll arr[1000010],oula[1000010];
/*ll euler(ll n)
{
ll res=n;
for(ll i=2;i*i<=n;++i)
{
if(n%i==0)
{
res=res/i*(i-1);
while(n%i==0)
n=n/i;
}
}
if(n>1)
res=res/n*(n-1);
return res;
}*/
void intt()
{
clean(arr,0);
clean(oula,0);
arr[0]=0;
arr[1]=0;
oula[1]=1;
for(int i=2;i<1000010;++i)
{
if(oula[i]==0)
{
for(int j=i;j<1000010;j=j+i)
{
if(oula[j]==0)
oula[j]=j;
oula[j]=oula[j]-oula[j]/i;
}
}
}
for(int i=2;i<1000010;++i)
arr[i]=arr[i-1]+oula[i];
}
int main()
{
intt();
int n;
while(cin>>n&&n!=0)
cout<<arr[n]<<endl;
}