算法设计：用黎曼和求π的近似值（含有python,C语言)

import numpy as npdef Riemann_pi(N):    pi = 0    fin_sum = 0    for i in range(0,N):        mid = (i+0.5)/N        fin_sum += 1/N * np.sqrt(1-mid**2)    pi = fin_sum*4    return piRiemann_pi(100000)

#include<stdio.h>int main(void){  float n,k;    double pi;
  // printf("please enter a 'n'\n");  scanf("%f",&n);  for(k=1,pi=0;k<=n;k++)  {    pi+=n/(n*n+k*k);  }  printf("pi=%f\n",4*pi);
  return 0;}