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. JDK安装及环境变量配置详解

    一.下载(Jdk 1.8 ) 1.下载地址:http://www.oracle.com/technetwork/java/javase/downloads/jdk8-downloads-2133151 ...

  2. django 数据库中中文转化为韩语拼音

    1.安装模块 django-uuslug pip install django-uuslug 2.导入模块 from uuslug import slugify 3.使用模块 slugify('天龙八 ...

  3. C# 替换去除HTML标记方法(正则表达式)

    [from] http://blog.csdn.net/sgear/article/details/6263848/// <summary> /// 将所有HTML标签替换成"& ...

  4. Spring中c3p0连接池 jar包下载 c3p0-0.9.2.1 jar包和mchange-commons-java-0.2.3.4 jar 包

    c3p0-0.9.2.1 jar包和mchange-commons-java-0.2.3.4 jar 包 下载地址: https://pan.baidu.com/s/1jHDiR7g 密码 tyek

  5. TCP/IP协议头部结构体

    TCP/IP协议头部结构体(转) 网络协议结构体定义 // i386 is little_endian. #ifndef LITTLE_ENDIAN #define LITTLE_ENDIAN (1) ...

  6. OpenCV3.42+VS2017配置+模块计算机类型“X86”与目标计算机类型“x64”冲突”的问题解决

    目录 OpenCV3.42+VS2017配置 Visual Studio 2017 第三方依赖设置,附加依赖项和附加库目录 "fatal error LNK1112: 模块计算机类型&quo ...

  7. HDU-1072-Nightmares

    这题可以用dfs写,我们记忆化搜索. 我们定义一个step和time数组,分别表示走到这点的最小步数,time表示走到该点炸弹还剩多少时间. 递归边界一是,如果走到该点,时间等于0,我们就返回. 如果 ...

  8. 【思维题 欧拉图】loj#10106. 单词游戏

    巧妙的模型转化 题目描述 来自 ICPC CERC 1999/2000,有改动. 有 NNN 个盘子,每个盘子上写着一个仅由小写字母组成的英文单词.你需要给这些盘子安排一个合适的顺序,使得相邻两个盘子 ...

  9. PHP实现同array_column一样的功能

    不用PHP自带的array_column函数实现同样的功能 <?php /** * Created by PhpStorm. * User: 123456 * Date: 2018/9/25 * ...

  10. NFS搭建

    一.环境 nfsserver01:192.168.127.100  centos7.3 nfsclient01:192.168.127.101  centos7.3 二.NFS原理 三.安装测试 1. ...