Fxtz IITime
limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total Submission (s): 525 Accepted Submission (s): 276
Problem Description
Cirno is playing a fighting game called "Fxtz" with Sanae.
Sanae is a Chushou (master) of the game while Cirno is a Shabao (noob). Since Cirno is a shabao, she just presses a random key on the keyboard for every 0.5 second, expecting to make a bishaji.
The battle begins. Having tried exactly 9 times, she finally makes a bishaji! She find herself summoned N iceballs!!! Then Sanae's HP decreases to 0 immediately .... It should has been like that. But Cirno was too simple and always navie. She doesn ' t know how to handle the iceballs, so she starts-press the keyboard at random, again.
Let's see how the Iceball damages. Each iceball have a fixed energy:the first ball's energy are 2^0, the second ball's energy are 2^1,..., and the n-th ball ' s Energy is 2^ (N-1). The damage caused by an iceball are equal to their energy. Cirno'll shoot N times. Since Cirno is pressing the keyboard at random, each time Cirno would choose exactly one iceball with equal possibility to Shoot out. Once shot out, the Iceball can ' t be chosen again. And even worse, the target may be either she opponent or herself, with equal possibility (50%). What's a big shabao she is. =_=
During shooting, once Cirno ' s HP is less than Sanae ' s, she'll lose the game. Otherwise, she wins.
You may assume Sanae do nothing while Cirno's shooting (all damages is caused by Cirno ' s iceball), and their original HP is both 2^n (No one would die in the middle of the battle unless Cirno's HP is less than Sanae ' s).
Here comes the question:can you calculate the possibility of Cirno ' s victory?
InputThe first line an integer C (c<=30), the number of test cases.
The only line contains one integer N (0<n<=500), indicating the number of iceballs.
Outputfor each case output a fraction, the possibility of Cirno ' s victory. The fraction must be reduced.
Sample Input
214
Sample Output
1/235/128
SourceThe 36th ACM/ICPC Asia Regional Beijing site--online Contest
Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=4043
The main topic: There are two people, a, B play games, A, a, a, b initial blood volume is 2^n, a total of N ball, the first ball of the attack is 2^ (i-1), A to choose N ball attack, each time randomly select an attack, select the ball can no longer use, attack with 1/2 probability attack to oneself, there is 1/2 probability , in the course of the game, once the blood amount of B is less than a, then a wins, after the election of N ball A or not win B wins, now the probability of B wins
Title Analysis: Consider the first item for 1, the geometric series of 2 for the a[i] = S[i-1] + 1,s[i] for the first and so we consider the position of the most attacking ball, get the recursive type:
PN = (1/n) * (*) * (p1 + p2 + p3 +). + pn-1), simplify to get PN = 2n-1/2n * pn-1, because n is larger, do it in large numbers
Import java.io.*;import java.math.*;import java.util.*;p ublic class Main {public static void Main (string[] args) { C1/>scanner in = new Scanner (system.in); int t = In.nextint (); for (int i = 0; i < T; i++) { BigInteger a = Biginteger.one; BigInteger B = biginteger.valueof (2); BigInteger n = In.nextbiginteger (); for (BigInteger j = biginteger.valueof (2), J.compareto (n) <= 0; j = j.add (Biginteger.one)) { a = A.multiply (J. Multiply (biginteger.valueof (2)). Subtract (Biginteger.one)); b = b.multiply (J.multiply (biginteger.valueof (2))); BigInteger tmp = A.GCD (b); A = A.divide (TMP); b = B.divide (TMP); } System.out.println (A + "/" + B);}}}
HDU 4043 Eliminate witches! (Calculate probability formula + large number)
