计算π的近似值VBπ=2*2^2 1*3

公式:π/4=1-1/3+1/5-1/7+...求值n=0sn=0an=1dountilABS(an)

#include<stdio.h>main(){ longi,a=1,b=3; doublep=1,t=1,t1=0; while(t-t1>=1e-8

你给的公式不对!应该是π/2＝2^2/1*3×4^2/3*5×.×（这个公式求的精度不高!代码如下：DimPI1AsDoubleDimPI2AsDoubleDimTempAsDoubleDimnAsD

PrivateSubCommand1_Click()DimSAsDouble,jcAsDouble,NAsIntegerjc=1N=Val(InputBox("请输入N"))Fori=1ToNjc=j

Private Sub Command1_Click()   '如果是题目要求用循环镶嵌个人觉得应该这样写   

#include#includevoidmain(){doublepi,sum=0,n=1;inti=1;while(n>=0.00001){//设置精度n=1/(float)(i*i);//计算各项

你好!PrivateSubForm_Click()DimiAsIntegerDimsAsIntegerFori=1To100s=s+iNextiPrintsEndSu

PrivateSubCommand1_Click()DimnAsInteger,iAsInteger,jAsInteger,Sum&,m&n=Val(InputBox("请输入N值"))Sum=0Fo

应应该为printf(

#includemain(){intn,i;doublet,sum;/*1*/printf("请输入n的值\n");scanf("%d",&n);sum=2;i=1;t=2;/*2*/while(i

DimsignAsInteger,aAsLong,piAsDoublesign=-1Fori=1To100000a=2*i-1sign=-1*signpi=pi+(1/a)*signNextpi=pi

Privatefunctione()Dime1,n,jcn=0e1=1jc=1Don=n+1jc=n*jce1=e1+1/jcLoopUntil1/jc

PrivateSubCommand1_Click()x=InputBox("请输入x：")s=1fact=1Doi=i+1fact=fact*it=x^i/facts=s+tLoopUntilt

PrivateSubForm_Click()Dimi&,PI#Doi=i+1PI=PI+(-1)^(i-1)/(2*i-1)LoopUntil1/iPI=PI*4Print"π=";PIEndSu

我看了一遍,1)的程序上没有问题我猜测float是精确到小数点后7位,同时在编译器中转化的时候会有精度损失因为现在的编译器都是将float型转化成为double型的来进行计算的pi=pi*a/b;这一

functionjiecheng(iasinteger)asdoubleforj=1toijiecheng*=inextendfunctiondime=0dimi=0doe+=1/jiecheng(i

num=m*(4/(2*n-1));改为num=m*(4.0/(2*n-1));试试再问：！！！！！哦哦，还要把m=pow((-1),(n+1));放到循环里面

PrivateSubCommand1_Click()DimNAsDouble,PiAsDoublePi=2ForN=1To1000Pi=Pi*(2*N)^2/((2*N-1)*(2*N+1))IfN=

'cosx=1-x^2/2!+x^4/4!.+x^(2n)/(2n)!我写的代码：Private Sub Command1_Click()   

PrivateSubCommand1_Click()Dimn,pi,ipi=2n=Val(InputBox("请输入一个数"))Fori=1Tonpi=pi*((2*i)^2/((2*i-1)*(2*