Hunters

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1588    Accepted Submission(s):
1194

Problem Description
Alice and Bob are the topmost hunters in the forest, so
no preys can escape from them. However, they both think that its hunting skill
is better than the other. So they need a match.
In their match, the targets
are two animals, a tiger and a wolf. They both know that the tiger is living in
the south of the forest and the wolf is living in the north of the forest. They
decide that the one who kills the tiger scores X points and who kills the wolf
scores Y points. If the one who kills both tiger and wolf scores X+Y
points.
Before the match starts, Alice is in the east of the forest and Bob
is in the west of the forest. When the match starts, Alice and Bob will choose
one of the preys as targets. Because they haven't known the other's choice,
maybe they choose the same target. There will be two situations:
(1) If they
choose different targets, they both are sure of killing their respective
targets.
(2) If they choose the same target, the probability of Alice killing
the target is P, and the probability of Bob killing it is 1-P. Then they will
hunt for the other prey, also the probability of Alice killing it is P and the
probability of Bob killing it is 1-P.
But Alice knows about Bob. She knows
that the probability of Bob choosing tiger as his first target is Q, and the
probability of choosing wolf is 1-Q. So that Alice can decide her first target
to make her expected score as high as possible.
 
Input
The first line of input contains an integer T
(1≤T≤10000), the number of test cases.
Then T test cases follow. Each test
case contains X, Y, P, Q in one line. X and Y are integers and 1≤X,
Y≤1000000000. P and Q are decimals and 0≤P, Q≤1, and there are at most two
digits after decimal point.
 
Output
For each test case, output the target Alice should
choose and the highest expected score she can get, in one line, separated by a
space. The expected score should be rounded to the fourth digit after decimal
point. It is guaranteed that Alice will have different expected score between
choosing tiger and wolf.
 
Sample Input
3
2 1 0.5 0.5
2 1 0 1
7 7 0.32 0.16
 
Sample Output
tiger 1.7500
wolf 1.0000
tiger 6.5968
题意:两个猎人比赛打猎,现在猎物有老虎和狼,猎杀老虎有X分,猎杀狼有Y分,并且其中一个猎人Alice知道同伴选择猎杀老虎的概率Q,猎杀狼的概率为(1-Q),并且如果两个人分别独自猎杀狼或者老虎,都能百分百将猎物猎杀,但如果他们选择
猎杀同一猎物,不管猎杀谁,Alice都有P的概率率先猎杀猎物,而其同伴则有(1-P)的概率猎杀猎物,求Alice的期望分数,并输出最高的期望分数(选狼还是选老虎)。
思路:概率问题,程序就是计算得到的期望公式就好了,没什么难度。期望公式
Alice选老虎的E(score)=( - Q)*X + Q*(X*P + Y*P) 
Alice选狼的   E(score)=Q*Y + ( - Q)*(Y*P + X*P)
AC代码:
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std;
const int N_MAX = +;
double P, Q,X,Y;
int main() {
int T;
scanf("%d",&T);
while (T--) {
scanf("%lf%lf%lf%lf", &X, &Y, &P, &Q);
double ext_1 = ( - Q)*X + Q*(X*P + Y*P);
double ext_2 = Q*Y + ( - Q)*(Y*P + X*P);
if (ext_1 > ext_2) {
printf("tiger %.4lf\n", ext_1);
}
else
printf("wolf %.4lf\n",ext_2);
}
return ;
}

poj 4438 Hunters的更多相关文章

  1. HDU 4438 Hunters

    Hunters Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Sub ...

  2. HDU 4438 Hunters (数学,概率计算)

    题意:猎人A和B要进行一场比赛.现在有两个猎物老虎和狼,打死老虎可以得X分,打死狼可以得Y分.现在有两种情况: (1)如果A与B的预定目标不同,那么他们都将猎到预定的目标. (2)如果A与B的预定目标 ...

  3. HDU 4438 Hunters 区域赛水题

    本文转载于 http://blog.csdn.net/major_zhang/article/details/52197538 2012天津区域赛最水之题: 题意容易读懂,然后就是分情况求出A得分的数 ...

  4. POJ 3370. Halloween treats 抽屉原理 / 鸽巢原理

    Halloween treats Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7644   Accepted: 2798 ...

  5. POJ 2356. Find a multiple 抽屉原理 / 鸽巢原理

    Find a multiple Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7192   Accepted: 3138   ...

  6. POJ 2965. The Pilots Brothers' refrigerator 枚举or爆搜or分治

    The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22286 ...

  7. POJ 1753. Flip Game 枚举or爆搜+位压缩,或者高斯消元法

    Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37427   Accepted: 16288 Descr ...

  8. POJ 3254. Corn Fields 状态压缩DP (入门级)

    Corn Fields Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9806   Accepted: 5185 Descr ...

  9. POJ 2739. Sum of Consecutive Prime Numbers

    Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20050 ...

随机推荐

  1. CornerStone使用跳坑总结

    Cornerstone是专门为Mac用户设计的Subversion(SVN)的控制,使版本控制更加透明.cornerstone根Xcode相比,能够更好的忽略文件,所以除了项目经理第一次初始化项目的时 ...

  2. Java写诗程序

    import java.util.Random; public class test_word { public static void main(String[] args) { System.ou ...

  3. iOS 解决ipv6问题

    解决ipv6的方法有很多种,由于现在国内的网络运营商还在使用ipv4的网络环境,所以appstore应用不可能大范围去修改自己的服务器, 而且国内的云服务器几乎没有ipv6地址. 这里附上苹果开发平台 ...

  4. 解决升级mac os X EI Capitan后遇到LibclangError: dlopen(libclang.dylib, 6): image not found.的问题

    打开文件./frameworks/cocos2d-x/tools/bindings-generator/clang/cindex.py 把第 3395 行 改为 : library = cdll.Lo ...

  5. c内置数据类型

    参考 C与指针 第三章 类型 类型标识符 字节 表示数值范围 备注 整型 [signed] int 2* -32768~32767 -2^15 ~ (2^15 -1) 无符号整型 unsigned [ ...

  6. 【linux】 服务器文件说明

    文件名 说明 /etc/resolv.conf  域名解析服务器地址文件 /etc/services 服务程序对应端口号文件 /etc/passwd 登录账号文件 /etc/hosts 本地IP域名解 ...

  7. LeetCode(224) Basic Calculator

    题目 Implement a basic calculator to evaluate a simple expression string. The expression string may co ...

  8. 数学基础:HDU2802-F(N)(寻找循环节)

    F(N) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submissi ...

  9. python 提交form-data之坑

    #coding=utf-8 import requests from requests_toolbelt import MultipartEncoder #requests库上传 files = {& ...

  10. [转] 在vue中使用SockJS实现webSocket通信

    最近接到一个业务需求,需要做一个聊天信息的实时展示的界面,这就需要和服务器端建立webSocket连接,从而实现数据的实时获取和视图的实时刷新.在此将我的实现记录下来,希望可以给有同样需求的人一些帮助 ...