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【规律】Farey Sums

时间:2019-08-21 11:43:18      阅读:69      评论:0      收藏:0      [点我收藏+]

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【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

参考博客:

 

别人博客的推导公式:

 

技术图片

 

 

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 

 


 

 

技术图片
 1 #pragma GCC optimize("Ofast,no-stack-protector")
 2 #pragma GCC optimize("O3")
 3 #pragma GCC optimize(2)
 4 #include <bits/stdc++.h>
 5 #define inf 0x3f3f3f3f
 6 #define linf 0x3f3f3f3f3f3f3f3fll
 7 #define pi acos(-1.0)
 8 #define nl "\n"
 9 #define pii pair<ll,ll>
10 #define ms(a,b) memset(a,b,sizeof(a))
11 #define FAST_IO ios::sync_with_stdio(NULL);cin.tie(NULL);cout.tie(NULL)
12 using namespace std;
13 typedef long long ll;
14 const int mod = 998244353;
15 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;}
16 //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;}
17 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;}
18  
19 const int N = 1e4+5;
20  
21 ll pki[N];
22  
23 void get_pki()
24 {
25     for(int i=1;i<N;i++) pki[i] = i;
26     for(int i=2;i<N;i++){
27         if(pki[i]==i)for(int j=i;j<N;j+=i)
28             pki[j]=pki[j]/i*(i-1);
29         pki[i] += pki[i-1];
30     }
31 }
32  
33 int main()
34 {
35     get_pki();
36     int _, cas, n;
37     for(scanf("%d",&_);_--;)
38     {
39         scanf("%d",&cas);
40         scanf("%d",&n);
41         printf("%d %lld/2\n",cas,3*pki[n]-1);
42  
43     }
44  
45 }
View Code

 

【规律】Farey Sums

标签:||   tps   bsp   stack   sdn   relative   span   参考   show   

原文地址:https://www.cnblogs.com/Osea/p/11387238.html

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