2011 Asia Beijing Regional Online Contest-1004 hdu4043 FXTZ II

很简单的数学题（其实主要是考高精度），枚举第一个打出的是什么雪球就可以找出递推式。

假设一共有n个雪球，如果第一次打出的是最小的，并且打在Sanae身上，那么赢的概率就是再打n-1个雪球赢的概率（若打在Cirno身上，他直接输），如果第一次打的是第k小的球，则赢的概率就应是再打n-k个球赢的概率（k前的球不会影响结果）

根据递推关系可得递推式为： dp[n] = 1.0/(2*n)*(dp[n-1]+dp[n-2]+dp[n-3]+......+dp[0])

dp[n-1] = 1.0/(2*n-2)*(dp[0]+dp[1]+...+dp[n-2]) 将dp[n-1]带入dp[n]中可得 dp[n]=(2*n-1)/(2*n)/dp[n-1]

直接用Java BigInteger就可以过了

import java.math.BigInteger;
import java.util.Scanner;public class HDOJ_1212 {public static void main(String[] str) {Integer num;BigInteger tmp;BigInteger[] fenzi = new BigInteger[505];BigInteger[] fenmu = new BigInteger[505];Scanner cin = new Scanner(System.in);int k, n;fenzi[0] = new BigInteger("1");fenmu[0] = new BigInteger("1");for (int i = 1; i <= 500; i++) {num = 2 * i - 1;tmp = new BigInteger(num.toString());fenzi[i] = fenzi[i - 1].multiply(tmp);num = 2 * i;tmp = new BigInteger(num.toString());fenmu[i] = fenmu[i - 1].multiply(tmp);tmp = gcd(fenzi[i], fenmu[i]);fenzi[i] = fenzi[i].divide(tmp);fenmu[i] = fenmu[i].divide(tmp);}while (cin.hasNext()) {k = cin.nextInt();for (int i = 0; i < k; i++) {n = cin.nextInt();System.out.println(fenzi[n].toString() + "/"+ fenmu[n].toString());}}}public static BigInteger gcd(BigInteger a, BigInteger b) {if (b.equals(new BigInteger("0")))return a;elsereturn gcd(b, a.mod(b));}
}

2011 Asia Beijing Regional Online Contest-1004 hdu4043 FXTZ II相关推荐

2018 ACM-ICPC Asia Beijing Regional Contest题解
以下所有AC题解程序来自"仙客传奇"团队. A. Jin Yong's Wukong Ranking List AC的C++语言程序: #include <iostream& ...
The 2020 ICPC Asia Shenyang Regional Programming Contest I题 Rise of Shadows（数论）
题目链接The 2020 ICPC Asia Shenyang Regional Programming Contest 题目大意: 一天内有H小时,每小时M分钟,时针分针以恒定速率旋转. 现在若时针 ...
The 2020 ICPC Asia Yinchuan Regional Programming Contest
The 2020 ICPC Asia Yinchuan Regional Programming Contest A 开三个vector数组存储x,y,z轴上的点,unique+erase去重 #in ...
The 2022 ICPC Asia Hangzhou Regional Programming Contest
比赛连接:Dashboard - The 2022 ICPC Asia Hangzhou Regional Programming Contest - Codeforces A. Modulo Rui ...
HDU 4059 The Boss on Mars （容斥）（2011 Asia Dalian Regional Contest）
The Boss on Mars 思路显然我们可以求得∑i=1ni4=6n5+15n4+10n3−n30\sum_{i = 1} ^{n} i ^ 4 = \frac{6n^5 + 15n^4 + ...
The 2021 ICPC Asia Taipei Regional Programming Contest
待补:C.Community Service/D.Largest Remainder A. Ice Cream 直接O(1)O(1)O(1)计算即可. #include <bits/stdc++ ...
2021 ICPC Asia Taipei Regional Programming Contest C、F
C.Community Service 题意: 有一个0到1e60到1e60到1e6的数轴,有2e52e52e5次操作: 操作一:增加一条从l到rl到rl到r的线段,每条线段有名字: 操作二:给定一 ...
The 2020 ICPC Asia Yinchuan Regional Programming Contest 银川 B. The Great Wall 题解
题目链接:Problem - B - Codeforces input1: 5 1 2 3 4 5 output1: 4 3 2 1 0 input2: 5 1 2 1 2 1 output2: 1 ...
The 2021 ICPC Asia Shanghai Regional Programming Contest 2021ICPC上海站VP
PS:实际打了2两个小时左右,被叫走去干活了.所以就直接补题了 A B C D E F G H I J K L M – – 补 AC AC – AC AC AC 补补 – 补 C.Strange M ...

2011 Asia Beijing Regional Online Contest-1004 hdu4043 FXTZ II

2011 Asia Beijing Regional Online Contest-1004 hdu4043 FXTZ II相关推荐

最新文章

热门文章