UC知道英文翻译，最好加上java程序
来源：百度知道 编辑：UC知道 时间：2024/06/02 09:08:50
这是一个java编程题，题目是英文的，想要翻译，如果能完成程序编写，追加100分
Pi can be calculated using the following sum, which is an approximation to Pi/4 :
PiSum = 1 - 1/3 + 1/5 - 1/7 + ... + (-1)^n / (2*n + 1)
1) Write two methods that return PiSum for a specified value of n, one method
using recursion and the other using iteration to do the sum.
Hint: instead of calculating (-1)^n, you can check whether the nth term is
negative by checking whether n%2 (i.e. n modulo 2) is 1.
Assume that the methods are always called with a non-negative value of n, so you
don't need to worry about dealing with invalid parameter values.
2) Show the additional code you would add to the above methods to throw an
exception if n is less than zero (you don't need to rewrite the whole method). The
exception handler should generate an error message saying that the number of terms
in the sum must not be negative.
Write a code fragment (i.e. it doesn't need to be a co
Pi can be calculated using the following sum, which is an approximation to Pi/4 :
PiSum = 1 - 1/3 + 1/5 - 1/7 + ... + (-1)^n / (2*n + 1)
1) Write two methods that return PiSum for a specified value of n, one method
using recursion and the other using iteration to do the sum.
Hint: instead of calculating (-1)^n, you can check whether the nth term is
negative by checking whether n%2 (i.e. n modulo 2) is 1.
Assume that the methods are always called with a non-negative value of n, so you
don't need to worry about dealing with invalid parameter values.
2) Show the additional code you would add to the above methods to throw an
exception if n is less than zero (you don't need to rewrite the whole method). The
exception handler should generate an error message saying that the number of terms
in the sum must not be negative.
Write a code fragment (i.e. it doesn't need to be a co
我有一个类似的代码。。将就着用把
大概意思是一样的只是没有异常处理
public class a2and5
{
public static void main(String [] agrs)
{
int count=0;
double pi=0.0,l=1.0,diff,pi1;
diff=1;
while(Math.abs(diff)>0.01)
{
pi1=pi;
pi+=((4.0/(count*2.0+1.0))*l);
l*=-1.0;
System.out.print(pi+" "+count+"\n");
count++;
diff=pi1-pi;
}
}
}
/*
3.1467739659661436 192
3.136438048653482 193
3.1467208250030967 194
3.1364906459749635 195
3.1466687630233094 196
3.1365421807448284 197
3.146617747495458 198
3.136592684838816 199
3.1465677471829556 200 */
你试用期有100分？？？
圆周率可以计算出利用以下的总和,是一种近似Pi / 4。
PiSum = 1 - 1 / 3 + 1 -七+……+ 10 ^(- 1)n /(2 * n + 1)。
1)写的两种方法PiSum返回指定的值的方法,一个
用递归与其他采用迭代来做