题解 | #牛牛学数列3#

#include <stdio.h>

int main() {
    int n,count=0;
    double num1=0,num2=0,a=0,b=0;
    scanf("%d",&n);
    for(double j=1;j<=n;j++){
        count++;//用于判断奇偶
      if(count%2==0){
        num1+=-(1/j);//奇数的和
      }
      else if(count%2!=0){
        num2+=1/j;
      }
    }
    printf("%.3f",num1+num2);
    return 0;
}