HDU 4043 Eliminate Witches! (求概率推公式 ＋ 大数)

FXTZ II

Time 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 have been like that. But Cirno is too simple, always navie. She doesn't know how to handle the iceballs, so she starts to press the keyboard at random, again.
Let's see how the iceball damages. Each iceball has a fixed energy: the first ball's energy is 2^0, the second ball's energy is 2^1,..., and the N-th ball's energy is 2^(N-1). The damage caused by an iceball is equal to its energy. Cirno will shoot N times. Since Cirno is pressing the keyboard at random, each time Cirno will 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 her opponent or herself, with equal possibility(50%). What a big ShaBao she is. =_=
During shooting, once Cirno's HP is less than Sanae's, she will lose the game. Otherwise, she wins.
You may assume Sanae did nothing while Cirno's shooting(all damages are caused by Cirno's iceball), and their original HP are both 2^N (No one will 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?

 

Input

The first line an integer C (C<=30), the number of test cases.
For each case, the only line contains one integer N(0<N<=500), indicating the number of iceballs.

 

Output

For each case output a fraction, the possibility of Cirno's victory. The fraction must be reduced.

 

Sample Input

214

 

Sample Output

1/235/128

 

Source

The 36th ACM/ICPC Asia Regional Beijing Site —— Online Contest

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4043

题目大意：有两个人A，B玩游戏，A，B初始血量都为2^n，一共有n个球，第i个球的攻击力为2^(i - 1)，A要选n个球攻击，每次随机选择一个攻击B，选择完该球就不能再用，攻击时有1/2概率攻击到自己，有1/2概率攻击到B，游戏过程中，一旦B的血量小于A，则A胜，当选完n个球A还是未胜则B胜，现在求B胜的概率

题目分析：考虑首项为1，公比为2的等比数列的第i项a[i] = s[i - 1] ＋ 1，s[i]代表前i项和，所以我们考虑攻击力最大的那个球的位置，得到递推式：
pn = (1 / n) * (1 / 2) * (p1 + p2 + p3 +..+ pn-1)，化简得到pn = 2n-1 / 2n * pn -1，因为n比较大，用大数做

import java.io.*;import java.math.*;import java.util.*;public class Main {    public static void main(String[] args){        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);        }    }}