#include "stdafx.h"
#include "stdio.h"
#include "math.h"

double pai(int n)
{
    double mult1 = 1,mult2 = 2,sum = 0.5;
    int index1,index2,index;
    if(n==1)
     ;
    else

        for(index = 2;index <= n;index++)
        {
              for(index1 = 1;index1 <= (2 * index - 3);index1 += 2)
              {
                mult1 *= index1;   
              }

                for(index2 = 1;index2 <= (2 * index - 2);index2 += 2)
                {
                    mult2 *= index2;   
                }
        sum += ((mult1/mult2/(2 * index - 1))*pow(0.5,2 * n - 1));
         
        }

return sum*6;

}

int main(int argc, char* argv[])
{
    int n;  
    int N=1;
    printf("请输入项数：");

    while(scanf("%d",&n)==0||n<=0)
    {
        printf("请输入大于零的整数!!!\n");
    }
    printf("项数N\t\tπ的估计值\n");
   
    while(N<=n)
    {
        printf("%d\t\t%f\n",N,pai(N));
        N++;
   
    }
   
    return 0;
}

w

此类问题你首先要描述的是  你使用哪个公式来取π的近似值

                 第1项输出0.5
                 第n（n>=2）项的通项为(1*3*5*7*…..*(2n-3))/(2*4*6*8*….*(2n-2))/(2n-1)*0.5^(2n-1)



double pai(int n)
{
    double mult1 = 1,mult2 = 2,sum = 0.5;
    int index1,index2,index;
    if(n==1)
     ;
    else

        for(index = 2;index <= n;index++)
        {
              for(index1 = 1;index1 <= (2 * index - 3);index1 += 2)
              {
                mult1 *= index1;   
              }

                for(index2 = 1;index2 <= (2 * index - 2);index2 += 2)
                {
                    mult2 *= index2;   
                }
        sum += ((mult1/mult2/(2 * index - 1))*pow(0.5,2 * n - 1));
         
        }

return sum*6;
用这部分函数来实现

这是啥公式  没看懂

比如  当=10时

（1*3*5*7*9*11*13*15*17）/（2*4*6*8*10*12*14*16*18）/19*0.5^19  这东西能求出π来  ？？？

是这样的sum += ((mult1/mult2/(2 * index - 1))*pow(0.5,2 * n - 1));用这个去累加

是这样的sum += ((mult1/mult2/(2 * index - 1))*pow(0.5,2 * n - 1));用这个去累加

这个公式太过于繁琐了  

#include<stdio.h>

double mypow(double x,int n)
{
    if(n==0) return 1.0;
    double t=1.0;
    for(;n>0;t*=x,n--);
    return t;
}

int main()
{
   int testn=20;
    int i,j;
    double t=1.0;
    double s=0.0;
    for(i=0;i<=9;i++)
    {
        t=1.0;
        for(j=1;j<=i*2+1;j+=2)
        {
            t*=(1.0*j)/(1.0*j+1);
        }
        t*=1.0/(i*2+3);
        t*=mypow(0.5,i*2+3);
        s+=t;
    }
    double pi=6.0*(0.5+s);
    printf("PI=%.8f\n",pi);
    return 0;
}