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 Node 35 { 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)const 48 { 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)const 57 { 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 }