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  • 近似计算

    #include<stdio.h>                                                     
    int main()                                                            
    {                                                                     
     double s=0, si,i;                                                    
     int n;                                                               
        for(int i=0;;i++)                                                 
     {                                                                    
      double term=1.0/(i*2+1);if(i%2==0)                                  
       if(i%2==0)                                                         
        s+=term;                                                          
       else s-=term;                                                      
       if(term<1e-6)break;                                                
      }                                                                   
      printf("%.6lf ",s);                                                
      return 0;                                                           
    }                                                                     
                                                                          

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  • 原文地址:https://www.cnblogs.com/freemain/p/4140661.html
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