break 因子 blank using one %d seq color rms

題目鏈接：http://poj.org/problem?id=2478

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19736		Accepted: 7962

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

Source

POJ Contest,Author:[email protected] 題目大意：很容易可以發現是求2-n的所有數的歐拉函數值之和 看代碼：

/**
有三條特性
若a為質數  phi[a]=a-1
若a為質數,b%a==0   phi[a*b]=phi[b]*a;
若a b 互質  phi[a*b]=phi[a]*phi[b]（當a為質數  如果b%a!=0）

 
*/
#include<iostream>
#include<cstdio>
using namespace std;
typedef long long LL;
const int maxn=1e6+50;
int phi[maxn],prime[maxn],p[maxn];//phi[i]代表i的歐拉函數值  prime[i]=0代表是素數 1代表不是素數   p存儲素數
void make()
{
    phi[1]=1;//特例
    int num=0;
    for(int i=2;i<=maxn;i++)
    {
        if(!prime[i])//是素數
        {
            p[num++]=i;//
            phi[i]=i-1;//素數的歐拉函數值就是它的值減1
        }
        for(int j=0;j<num&&p[j]*i<maxn;j++)//用當前已經得到的素數篩去p[j]*i
        {
            prime[p[j]*i]=1;//可以確定p[j]*i不是質數
            if(i%p[j]==0)//第二條特性
            {
                phi[p[j]*i]=phi[i]*p[j];
                break;//歐拉篩的核心語句  保證每個數只會被自己最小的質因子篩掉一次
            }
            else phi[p[j]*i]=phi[i]*phi[p[j]];
        }
    }
    return ;
}

int main()
{
    make();
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0) break;
        LL sum=0;
        for(int i=2;i<=n;i++) sum+=phi[i];
        printf("%lld\n",sum);
    }
//    for(int i=1;i<=100;i++) cout<<phi[i]<<" ";

    return 0;
}

Farey Sequence（歐拉函數板子題）