【规律】Farey Sums

【参考博客】：

https://blog.csdn.net/meopass/article/details/82952087

Farey Sums

题目描述

Given a positive integer, N, the sequence of all fractions a/b with (0 < a ≤ b), (1 < b ≤ N) and a and b relatively prime, listed in increasing order, is called the Farey Sequence of order N.
For example, the Farey Sequence of order 6 is:

0/1, 1/6, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 5/6, 1/1

If the denominators of the Farey Sequence of order N are:

b[1], b[2], . . . , b[K]

then the Farey Sum of order N is the sum of b[i]/b[i + 1] from i = 1 to K—1.
For example, the Farey Sum of order 6 is:

1/6 + 6/5 + 5/4 + 4/3 + 3/5 + 5/2 + 2/5 + 5/3 + 3/4 + 4/5 + 5/6 + 6/1 = 35/2

Write a program to compute the Farey Sum of order N (input).

输入

The first line of input contains a single integer P, (1 ≤ P ≤ 10000), which is the number of data sets that follow. Each data set should be processed identically and independently.
Each data set consists of a single line of input. It contains the data set number, K, followed by the order N, (2 ≤ N ≤ 10000), of the Farey Sum that is to be computed.

输出

For each data set there is a single line of output. The single output line consists of the data set number,K, followed by a single space followed by the Farey Sum as a decimal fraction in lowest terms. If the denominator is 1, print only the numerator.

样例输入

4
1 6
2 15
3 57
4 9999

样例输出

1 35/2
2 215/2
3 2999/2
4 91180457/2

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参考博客:

别人博客的推导公式:

我倒是觉得，这个题目就是找规律。因为看到答案都是分母为2，很容易想到其实这个题目就是找规律有关的。

联系互素就想到欧拉函数。写出前几个出来发现就是  ( 3phi(x) -  1 ) / 2

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 #pragma GCC optimize("Ofast,no-stack-protector")
 #pragma GCC optimize("O3")
 #pragma GCC optimize(2)
 #include <bits/stdc++.h>
 #define inf 0x3f3f3f3f
 #define linf 0x3f3f3f3f3f3f3f3fll
 #define pi acos(-1.0)
 #define nl "\n"
 #define pii pair<ll,ll>
 #define ms(a,b) memset(a,b,sizeof(a))
 #define FAST_IO ios::sync_with_stdio(NULL);cin.tie(NULL);cout.tie(NULL)
 using namespace std;
 typedef long long ll;
 const int mod = ;
 ll qpow(ll x, ll y){ll s=;while(y){if(y&)s=s*x%mod;x=x*x%mod;y>>=;}return s;}
 //ll qpow(ll a, ll b){ll s=1;while(b>0){if(b%2==1)s=s*a;a=a*a;b=b>>1;}return s;}
 inline int read(){int x=,f=;char ch=getchar();while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}while(ch>=''&&ch<='') x=x*+ch-'',ch=getchar();return x*f;}

 const int N = 1e4+;

 ll pki[N];

 void get_pki()
 {
     for(int i=;i<N;i++) pki[i] = i;
     for(int i=;i<N;i++){
         if(pki[i]==i)for(int j=i;j<N;j+=i)
             pki[j]=pki[j]/i*(i-);
         pki[i] += pki[i-];
     }
 }

 int main()
 {
     get_pki();
     int _, cas, n;
     for(scanf("%d",&_);_--;)
     {
         scanf("%d",&cas);
         scanf("%d",&n);
         printf("%d %lld/2\n",cas,*pki[n]-);

     }

 }