如何利用循环相加结局问题

int n,x,y;
    long count;
    double pi;
    printf("请输入一个整数\n");
    scanf("%d",&n);
    for(x=-n,count=0;x<n;x++)
        for(y=-n;y<n;y++)
            if((x*x+y*y)<n*n)
                count++;
    pi=count/(double)n/(double)n;
    printf("圆周率为%lf",pi);

#include <iostream>
using namespace std;
int main()
{
      double d =0,flag = 1;
      int i=1;
      while(1.0/i>=1e-6)
      {
            d += flag /i;
            flag = flag *-1;
            i+=2;
      }
      cout<<4*d<<endl;
      return 0;
}

代码如下：
记得采纳

double func_pai(int i)
{
    double init = 1;
    int flag = 1;
    double sum = 0;
    
    while(i--){
        sum = sum + flag/init;
        flag *= -1;
        init += 2;
    }
    return 4*sum;
}

#include<stdio.h>
#include<math.h>
int main(void)
{
int n;
long N;
double i, a, pi;
scanf_s("%d", &N); 、
n = 1;
a = 0;
i = 1;
do
{
a = a - pow(-1,n)*i;
n = n + 1;
i = 1/(2 * (double)n - 1);
} while (n < N);
pi = 4 * a;
printf("圆周率的近似值为%.*f\n",n,pi); //n控制精度，实际能输出的精度有限。根据需要改写这个值
return 0;
}