2011．多项式求和

#include<stdio.h>

int main(){

         int a=1,b=1,n,i;

         int arr[]={1};

         float s=arr[0];

         scanf("%d",&n);

         for(i=1;i<n;i++){

                   b+=1;

                   arr[i]=a/b;

                   if(i%2!=0) arr[i]=-arr[i];

                   else       arr[i]=arr[i];

                   s+=arr[i];

         }

         printf("%.2f",s);

         return 0;

}

2011．多项式求和

原文：https://www.cnblogs.com/huoyuying/p/9735687.html