Bnuoj34976 count Bacteria

Count Bacteria Time limit:1000 ms Memory limit:65536 kbspecial Judge64-bit integer Io format: % LLD Java class name: mainprev submit status Statistics discuss nextfont size: +  -

James discovered a bacteria that will be re-split into three bacteria on the first day after the split, that is, if one newly split bacteria exists on the first day, there is still one bacteria in the first day, 3 new bacteria split out in the second day, 3 bacteria in the second day, 9 bacteria in the second day... It is known that there are several of these bacteria in a petri dish, 1st days there are a bacterial, 2nd days there are B bacteria. Tom wants to know how many bacteria are in the petri dish on the nth day. The result (ANS) may be very large and only logs are output._{10 (ANS.}

Input

The first input behavior is an integer T (T ≤ 1000) indicating the number of data groups.

The three numbers of data in each group are a, B, n (1 ≤ A, B, n ≤ 10 in sequence.^{18 ).}

Output

For each group of data, a floating point number is output to indicate the result. The error is within 1-6.

Sample Input

11 3 3

Sample output

0.4771212547

Source 12th Beijing Normal University Program Design Competition finals authortemperlsyer

# Include <stdio. h> # include <math. h> int main () {int t; long K1, K2, n, a, B; scanf ("% d", & T); While (t --) {scanf ("% LLD", & A, & B, & N); long TA = A, TB = B, mm; A = (Tb-Ta)/2; // number of cells in the first day after split. B = Ta-; // Number of new cells in the first day k1 = n/2; k2 = (n-1)/2; If (k1 = k2) Mm = 1; else Mm = 3; if (n> = 3) printf ("%. 10lf \ n ", K2 * log10 (3.0) + log10 (double) (A * mm + B ))); // The multiplication in the log equals to the addition of else if (n = 1) printf ("%. 10lf \ n ", log10 (double) (TA * 1.0); else if (n = 2) printf (" %. 10lf \ n ", log10 (double) (Tb * 1.0 )));}}

Bnuoj34976 count Bacteria