Question

Death Battle

Problem Description

In a crossover fantasy universe, Houin Kyoma is up in a battle against a powerful monster Nomu that can kill him in a single blow. However being a brilliant scientist Kyoma found a way to pause time for exactly M seconds. Each second, Kyoma attacks Nomu with certain power, which will reduce his health points by that exact power. Initially Nomu has H Health Points. Nomu dies when his Health Points reach 0. Normally Kyoma performs Normal Attack with power A. Besides from Kyoma’s brilliance, luck plays a major role in events of this universe. Kyoma’s Luck L is defined as probability of performing a super attack. A super attack increases power of Normal Attack by C. Given this information calculate and print the probability that Kyoma kills Nomu and survives. If Kyoma dies print “RIP”.

Constraints

0 < T <= 50

1 <= A, H, C, L1, L2 <= 1000

1 <= M <= 20.

L1<=L2

Input Format

First line is integer T denoting number of test cases.

Each test case consist of single line with space separated numbers A H L1 L2 M C. Where luck L is defined as L1/L2. Other numbers are, as described above.

Output

Print probability that Kyoma kills Nomu in form P1/P2 where P1<=P2 and gcd(P1,P2)=1. If impossible, print “RIP” without quotes.

Timeout

1

Explanation

Example 1

Input

2

10 33 7 10 3 2

10 999 7 10 3 2

Output

98/125

RIP

Answer 1

Complete code in C++:-

#include <bits/stdc++.h>

using namespace std;

// For finding the factorial of a number 'a'
long long int fact(long long int a)
{
   long long int x = 1;
   while(a > 0)
   {
       x *= a;
       a--;
   }
   return x;
}

// For finding gcd of two numbers 'a', 'b'
long long int gcd(long long int a, long long int b)
{
   if(b == 0)
       return a;
   return gcd(b, a%b);
}

// Main function
int main(void)
{
   long long int a, h, l1, l2, m, c, t, p1, p2;

   // Teking user input for test cases.
   cout << "Input" << '\n';
   cin >> t;
   while(t--)
   {
       p1 = 0;
       p2 = 0;

       // Taking user input to process for each test case.
       cin >> a >> h >> l1 >> l2 >> m >> c;
       long long int temp = ceil((double)(h - a*m)/c);
       if(temp <= 0)
       {
           cout << 1 << "/" << 1;
       }
       else if(temp > m)
       {
           cout << "RIP";
       }
       else
       {
           p2 = pow(l2, m);
           while(temp <= m)
           {
               long long int ways = fact(m)/(fact(temp)*fact(m-temp));
               p1 += ways*(pow(l1, temp)*pow((l2-l1), (m-temp)));
               temp++;
           }
           temp = gcd(p1, p2);
           cout << p1/temp << "/" << p2/temp;
       }
       cout << '\n';
   }
   return 0;
}

Output:-