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题目：求出1~1000之间的所有能被7整除的数，并计算和输出每5个的和
首先我们来看，这道题需要循环所有1-1000之间的数，于是就想到了for循环
for (int value = 1; value <= 1000; value++)
使用for对1~1000的数进行循环，也相同于：
int i = 1; while(i <= 1000) i++;
当然，使用for循环一行就可以解决，我们就使用for循环了；
接着考虑所有能被7整除的数，那必然是这个数与7求余等于0，那么接着嵌套if语句：
if (value % 7 == 0)
看到每五个输出一次和，一开始是想到写一个for语句循环五次之后跳出，但是行不通，于是想到用一个值来统计能被7整除的数出现的次数；
在for之前初始化count（用来计数）和sum（用来统计每五个数的和）：
int count = 0; int sum = 0;
继续在if嵌套里面，先对找到的数进行加和，再令计数器加一：
sum += value; Console.WriteLine(value); count++;
现在进行判断，如果计数的值等于5，那么输出加和，并且清空计数器和加和：
if (count == 5) { Console.WriteLine("这5个的和为：" + sum); sum = 0; count = 0; }
那么就可以输出最终结果：
源代码如下：
int count = 0; int sum = 0; for (int value = 1; value <= 1000; value++) { if
(value % 7 == 0) { sum += value; Console.WriteLine(value); count++; if (count
== 5) { Console.WriteLine("这5个的和为：" + sum); sum = 0; count = 0; } } }
将其写到Main方法下面即可
制作不易，求个一键三连~
