Question

哪位帮我写下程序 C语言 π/4=1-1/3+1/5-1/7+…+1/4n-3-1/4n-1+… 要求n=10000 求π的值 感激不尽

Answer 1

//第一种方法
#include<iostream.h>
#include<math.h>
void main()
{
double Pi=0;
double i=1;
for(double n=1;n<=1000000;n+=2)
{
Pi+=(pow(-1,(i+1))*(1/n));
i++;
}
cout<<4*Pi<<endl;
}

//第2种方法,自己构建pow()函数
#include<iostream.h>
double funcc( int a);
void main()
{
unsigned long double Pi=1;
int i=2;
for(unsigned long double n=3;n<999999;n+=2)//这里n一定要是double，n的运行范围决定了π的精度
{
Pi+=((funcc(i))*(1/n));//否则这里1/n的值是0!!
//cout<<funcc(i)<<" "<<Pi<<" "<<i<<endl;//调试
i++;
}
cout<<Pi*4<<endl;
}

double funcc(int a)
{
if(a%2==0)
return -1;
return 1;
}
///////////////////////////////////
c++写的..

Answer 2

int i=1;
int sign=-1;
double pi=1.0;
while(i<=10000) {
pi+=sign/(4*i-1);
i++;
sign*=-1;
}

pi*=4;

Answer 3

double sum1=0.0,sum2=0.0;
double pi=0.0;
for(int i = 1;i<10000;i++){
sum1=(double)(1.0/(4.0*i-3.0))+sum1;
sum2=(double)(1.0/(4.0*i-1.0))+sum2;
}
pi=4.0*(sum1-sum2);

Answer 4

#include<stdio.h>
int f(int n)
{
float TT;
if(n=1)
TT/4=1-1/3;
if(n>1)
TT/4=1/(4n-3)-1/(4n-1)+f(n-1);
return TT;
}
int main()
{
double s;
s=f(10000);
printf("%f",s);
}

Answer 5

#include "stdio.h"
void main()
{
int s;
float t,pi,n;
t=1;pi=0;n=1.0,s=1;
while(n<=10000)
{ pi=pi+t;
n=n+2;
s=-s;
t=s/n;
}
pi=pi*4;
printf("pi=%10.6f\n",pi);
}