【偶数求和】 （C语言代码）

解题思路:





注意事项:





参考代码:

#include <stdio.h>

int main()

{

    int  a[100];

    int i,j,k,n,m,sum1,sum2,x;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        sum1=0;

          sum2=0;

        j=1;

        k=n/m;

        for(i=1;i<=n;i++)

        {

          a[i]=2*i;

        }

        if(n==m)

        {

            printf("%d ",n+1);//等差数列求n项和公式，再除以n

        }

        if(n>m)

        {

            for(i=0;i<k;i++)

           {

               for(x=1;x<=m;x++)

             {

                sum1+=a[j];

                j++;

              }

                sum1=sum1/m;

                printf("%d ",sum1);

                sum1=0;//此步骤关键，记得将sum1初始为0

            }

        }

        if(n%m!=0)//不足m项的累加再求平均值

        {

            for(i=0;i<n%m;i++)

            {

                sum2+=a[j];

                j++;

            }

            sum2=sum2/(n%m);

            printf("%d ",sum2);

        }

        printf("\n");

    }

    return 0;

}