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POJ总结(3299)

标签:   poj  c  3299

     POJ总结(3299)POJ3299 Humidex题目思路:1.题目中涉及到三个变量,要求给出其中两个,计算出第三个,然后将三个内容都输出。其中涉及到的主要问题为:判断需要计算哪个参数根据上述判断执行计算过程2.首先解决第一个...

POJ3299

标签:   刷题

     POJ3299题解 POJ3299 解题重点在如下几个地方 第一:正确接收来自输入流的数据 第二:将数学公式正确翻译成C语言公式 第三:题目意思是任给两个变量求第三个,而不是像输入样例那样只给你T、D SOURCE 代码...

POJ-3299解题

标签:   POJ 3299

     POJ-3299解题 C++源代码 Description Adapted from Wikipedia, the free encyclopedia The humidex is a measurement used by Canadian meteorologists to reflect the combined effect of heat and humidity. It...

     #include<iostream> #include<cmath> #define exp 2.718281828 using namespace std; float geth(float t,... e=6.11*pow(exp,5417.7530*((1/273.16)-(1/(d+273.16))));... h=0.5555*...

poj 3299

标签:   poj 3299

     //poj 3299 数学公式 一次AC 但是写的比较多。。 #include #include #include #include using namespace std; int main() { char c; cin>>c; while(c != 'E') { double a, b; char c2; cin>>a; cin...

     Adapted from Wikipedia, the free encyclopedia The humidex is a measurement used by Canadian meteorologists to reflect the combined effect of heat and humidity. It differs from the heat index used in ...

     #include #include #include using namespace std; double HH(double TT,double DD) { double h,e; e=6.11*exp(5417.7530*((1/273.16)-(1/(DD+273.16)))); h=0.5555*(e-10.0); return TT+h;...

     Humidex Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19054   Accepted: 6885 Description Adapted from Wikipedia, the free ...The humidex is a me

      Humidex Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 24416   Accepted: 8639 Description ... Adapted from Wikipedia, the free encyclopedia...

     POJ3299 -- Humidex Submit1:Wrong Answer #include #include int main(){ char flag; double temperature,dew_point; while(scanf("%c",&flag)!=EOF&&flag!='E'){ scanf(" %lf D %lf",&t

     题目大意:有两个数,求第三个数。 题目:三个变量temperature,dewpoint,humidex有以下关系: humidex = temperature + h h = (0.5555)× (e - 10.0) e = 6.11 × exp [5417.7530 × ((1/273.16) - (1/(dewpoint...

POJ 3299 java

标签:   java

     import java.util.Scanner; public class Main { static double exp=2.718281828; public static void main(String[] args) { Scanner sc=new Scanner(System.in); while(true) { //初始化三个变量 ...

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