codeforce Round On Sum of Fractions

D. On Sum of Fractions

 

Let's assume that

• v(n) is the largest prime number, that does not exceed n;
• u(n) is the smallest prime number strictly greater than n.

Find .

Input

The first line contains integer t (1 ≤ t ≤ 500) — the number of testscases.

Each of the following t lines of the input contains integer n (2 ≤ n ≤ 109).

Output

Print t lines: the i-th of them must contain the answer to the i-th test as an irreducible fraction "p/q", where p, q are integers, q > 0.

Sample test(s)

input

2
2
3

output:1/6

7/30


1：欧几里得算法 计算GCD递归
gcd(a,b) == gcd(b,a%b);     a,b>0
2：判断素数:
3：符号重载

 1 #include<iostream> 2 #include<stdio.h> 3 #include<string> 4 #include<string.h> 5 #include<map> 6 #include<math.h> 7  8 using namespace std; 9 typedef long long LL;10 bool Is_Prime(LL x)11 {12     for(LL i=2;i*i<=x;i++)13         if(x%i == 0)14             return 0;15     return 1;16 }17 LL V_Down(LL x)18 {19     while(!Is_Prime(x))20         x--;21     return x;22 }23 LL U_Up(LL x)24 {25     x++;26     while(!Is_Prime(x))27         x++;28     return x;29 }30 LL gcd(LL x, LL y)31 {32     return y==0?x:gcd(y, x%y);33 }34 class Node35 {36     public:37         LL zi;38         LL mu;39     public:40         Node(){};41         Node(LL z ,LL m)42         {43             LL g=gcd(z,m);44             zi=z/g;45             mu=m/g;46         };47         Node operator +(const Node &A)const48         {49             LL m,z,g;50             g=gcd(mu,A.mu);51             m=mu/g*A.mu;52             z=A.mu/g*zi+mu/g*A.zi;53             g=gcd(z,m);54             return Node(z/g,m/g);55         }56         Node operator - (const Node & A)const57         {58             LL m,z,g;59             g=gcd(mu,A.mu);60             m=mu/g*A.mu;61             z=A.mu/g*zi-mu/g*A.zi;62             g=gcd(z,m);63             return Node(z/g,m/g);64         }65         Node &operator = (const Node &now)66         {67             this->mu=now.mu;68             this->zi=now.zi;69             return *this;70         }71 72         friend ostream & operator <<(ostream &os,const Node &A)73         {74             os<<A.zi<<"/"<<A.mu;75             return os;76         }77 78 79 };80 int main()81 {82     int t;83     LL x;84     cin>>t;85     while(t--)86     {87         cin>>x;88         LL v=V_Down(x);89         LL u=U_Up(x);90         Node ans=Node(1,2)-Node(1,v);91         Node sum=Node(x-v+1,u*v)+ans;92         cout<<sum<<endl;93     }94     return 0 ;95 }