【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 "
"
#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 = 998244353;
ll qpow(ll x, ll y){ll s=1;while(y){if(y&1)s=s*x%mod;x=x*x%mod;y>>=1;}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=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();return x*f;}
 
const int N = 1e4+5;
 
ll pki[N];
 
void get_pki()
{
    for(int i=1;i<N;i++) pki[i] = i;
    for(int i=2;i<N;i++){
        if(pki[i]==i)for(int j=i;j<N;j+=i)
            pki[j]=pki[j]/i*(i-1);
        pki[i] += pki[i-1];
    }
}
 
int main()
{
    get_pki();
    int _, cas, n;
    for(scanf("%d",&_);_--;)
    {
        scanf("%d",&cas);
        scanf("%d",&n);
        printf("%d %lld/2
",cas,3*pki[n]-1);
 
    }
 
}