c++程序设计设s=1+1/2+1/3+....+1/n,求与8最接近的s与n的值

如题所述

#include<stdio.h>

int main()

{ int i=0;

  double s=0;

  while(s<8)

    s+=1.0/++i;

  if(s-8<8-(s-1.0/(i-1)))

    printf("s=%f\ti=%d\n",s,i);

  else printf("s=%f\ti=%d\n",s-1.0/(i-1),i-1);

  return 0;

}