偶数求和

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 91317    Accepted Submission(s): 38494

Problem Description

有一个长度为n(n<=100)的数列，该数列定义为从2开始的递增有序偶数，现在要求你按照顺序每m个数求出一个平均值，如果最后不足m个，则以实际数量求平均值。编程输出该平均值序列。

Input

输入数据有多组，每组占一行，包含两个正整数n和m，n和m的含义如上所述。

Output

对于每组输入数据，输出一个平均值序列，每组输出占一行。

Sample Input

3 2

4 2

Sample Output

3 6

3 7

Author

lcy

Source

C语言程序设计练习（三）

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两种方法：

1：按正常思路来

#include <stdio.h>
int main()
{
    int n, m, i, sum, num, num2, j, s,s1;
    while (scanf_s("%d%d", &n, &m) != EOF)
    {
        sum = 0;
        int a[10001];
        s1=num = n / m;
        s = num2 = n%m;
        j = 0;
        for (i = 2; i <= n * 2; i += 2, j++)
            a[j] = i;
        j = 0;
        while (num--)
        {
            for (i = 0; i < m; i++)
            {
                sum += a[j];
                j++;
            }
            printf("%d", sum / m);
            if (num>0)
                printf(" ");
            sum = 0;
        }
        if (num2 == 0)
            printf("\n");
        else
        {
            while (num2--)
            {
                sum += a[j];
                j++;
            }
            if (sum != 0 && s == 0)
                printf(" %d\n", sum / (s + 1));
            if (sum != 0 && s != 0)
                printf(" %d\n", sum / s);
        }
    }
    return 0;
}2：简单很多，主要我也不知道我咋想到的！！！

#include <stdio.h>
int main(void)
{
    int a, i, n, m, b, s;
    while (scanf("%d %d", &n, &m) != EOF && n<=100)
    {
        a = 2;
        b = n%m;
        s = n / m;
        for (i = 0; i < s; i++)
        {
            printf("%d", a + m - 1);
            if (i < s - 1)
                printf(" ");
            a = a + m * 2;
        }
        if (b != 0)
        {
            printf(" %d", a);
        }
        printf("\n");
    }
    return 0;
}