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POJ 2478-Farey Sequence(篩選法求歐拉函數)

編輯：C++入門知識

POJ 2478-Farey Sequence(篩選法求歐拉函數)

Farey Sequence Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2478 Appoint description:

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

題意：給定一個數n，求小於或等於n的數中兩兩互質組成的真分數的個數。

思路：這個博客有關於這道題的推理---->點擊打開鏈接

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
using namespace std;
typedef long long LL;
const int inf=0x3f3f3f3f;
const double pi= acos(-1.0);
LL phi[1000010];
LL res[1000010];
void Euler()
{
    int i,j;
    memset(phi,0,sizeof(phi));
    phi[1]=1;
    for(i=2;i<=1000010;i++)
    {
        if(!phi[i])
        {
            for(j=i;j<=1000010;j+=i)
            {
                if(!phi[j]) 
                    phi[j]=j;
                phi[j]=phi[j]/i*(i-1);
            }
        }
    }
}
int main()
{
    int n;
    Euler();
    memset(res,0,sizeof(res));
    res[1]=res[2]=1;
    for(int i=3;i<1000010;i++)
        res[i]=res[i-1]+phi[i];
    while(~scanf("%d",&n)){
        if(!n) break;
        printf("%lld\n",res[n]);
    }
    return 0;
}