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  1.  
  2. #include <stdio.h>
  3. #include <math.h>
  4. int main() {
  5.     double p=0.0,x;
  6.     int n;
  7.     x =pow(-1,n)/ (2 * n + 1);
  8.     for(n=0;fabs(x)<=pow(10,-6);n++){
  9.         p+=x;
  10. }
  11.     p *= 4;
  12.     printf("π 的近似值为：%.8f\n", p);
  13.     return 0;
  14. }
Success #stdin #stdout 0.01s 5316KB
comments (?)
stdin 
Standard input is empty
stdout 
π 的近似值为：0.00000000
https://ideone.com/ENlALE
language:
C (gcc 8.3)
created:
208 days ago
visibility:
public

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