// Arup Guha
// 8/10/06
// Solution to 2006 UCF Local Contest Problem: Etch

// Adapted to C on 4/9/2010

#include <stdio.h>

#define EPSILON 0.0000001

double findTime(double f1, double f2, double a, double b, double c);
double f(double a, double b, double c, double t);

int main() {

    FILE *ifp;
    ifp = fopen("etch.txt", "r");
    int numcases;
    fscanf(ifp, "%d", &numcases);

    int i;

	// Loop through all of the input cases.
	for (i=1; i<=numcases; i++) {

	    // Read in the data for the case.
	    double f1, f2, a, b, c;
	    fscanf(ifp, "%lf%lf%lf%lf%lf", &f1, &f2, &a, &b, &c);

	    // Find the answer and print it out properly formatted.
	    double ans = findTime(f1, f2, a, b, c);
        printf("Crystal #%d: %.2lf\n\n",i,ans);
	}

	fclose(ifp); // Close the input file.

	system("PAUSE");
	return 0;
}

// This function doesn't analytically determine the correct time. Instead, since
// the function is monotonically increasing, it does a binary search until it
// gets pretty close to the actual value.
double findTime(double f1, double f2, double a, double b, double c) {

    // This is the value we want the function on the right-hand side to evaluate to.
    double targettime = (f2-f1)/(f1*f2);

    // The crystal has to etch for more than 0 seconds.
    double low = 0;

    // Based on the formula, since c is positive, we know that the amount of
    // etch time is less than if we divided the target by a, because this would
    // be the correct answer if we didn't take into account the contribution
    // to the function of the exponential part.
    double high = targettime/a;

    // Essentially, low represents a value lower than the actual etch time
    // and high represents a value greater than the actual etch time.

    // Assigning mid a value so the compiler doesn't yell.
    double mid = 0;

    // Keep on refining the guess until high and low are very close together.
    while (high - low > EPSILON) {

        // Guess in the middle of our boundary.
        mid = (high+low)/2;

        double ans = f(a,b,c,mid);

        // If this guess is too low, adjust our low bound.
        if (ans<targettime)
            low = mid;

        // Real lucky if your guess is perfect.
        else if (ans == targettime)
            break;

        // Otherwise adjust the high bound.
        else
            high = mid;
    }

    return mid; // This is the answer we want to return.
}

// Just calculates the function in the problem description
// in terms of a,b,c and t.
double f(double a, double b, double c, double t) {
    return a*t+b*(1-exp(-c*t));
}