poj2478——Farey Sequence（欧拉函数）

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18507		Accepted: 7429

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

题意：给出一个数n，要你求出在1-n中，互质的数有多少对；

思路：这题属于欧拉函数的应用。1-n中互质数的对数就是等于与2互质的数的个数+与3互质的数的个数+与4互质的数的个数+....+与n互质的数的个数。（为什么没有1，因为一个数无法成对）
所以这题就演变成了求2-n中所有数的欧拉函数的和，所以要用到欧拉函数的变形写法。
因为是多组输入，所以要预处理一下，提前记录下n为1-1000000时的值。

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
#define eps 1e-7
#define ll long long
#define inf 0x3f3f3f3f
#define pi 3.141592653589793238462643383279
using namespace std;
ll n,euler[1000006],sum[1000006];
int main()
{
    for(int i=1; i<=1000000; ++i) //初始化所有数的欧拉函数值为自己 
        euler[i] = i;
    for(int i=2; i<=1000000; ++i) //求所有数的欧拉函数值 
    {
        if(euler[i] == i) //根据公式 欧拉值（n） = n*（1-1/p1）（1-1/p2）...（1-1/pi）其中p为n的素因子
        //可知素数只有自己是自己的素因子，所以在遇到自己以前euler值没变，所以可以通过上面的语句得知euler[i]==i的为素数 
        {
            for(int j=i; j<=1000000; j+=i) //所有i的倍数乘上（1-1/i） 
                euler[j] = euler[j]/i*(i-1);
        }
    }
    sum[1] = 0;
    for(int i=2; i<=1000000; ++i) //提前记录下每个数的答案 
    {
        sum[i] = sum[i-1] + euler[i];
    }
    while(scanf("%lld",&n)!=EOF)
    {
        if(!n) break;
        printf("%lld\n",sum[n]);
    }
    return 0;
}