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题目：3个可乐瓶可以换一瓶可乐，现在有364瓶可乐。
问：一共可以喝多少瓶可乐，剩下几个空瓶！
首先先看题目 这是道数学题
（好家伙 真没想到还有数学问题qaq）
现在拿出草稿纸一起写：
3 个可乐瓶可以换 1 瓶可乐，
那么 364 瓶可乐就可以换 364/3 瓶可乐（由于/两边都是整数，所以得到的结果一定也是整数），
剩下 364%3 瓶可乐换不了，因为不满3瓶了嘛~
所以第一波，
喝的总数就是：（喝的+换的）
364 + 364/3
364 +121 =485
剩下的空瓶就有：（换的+没够换剩下的）
364/3 + 364%3
121+ 1 = 122
再来第二波，
喝的总数就是：（第一波喝的+第一波剩下的空瓶/3）
（364 + 364/3）+（364/3 + 364%3）/ 3
485 + 122 / 3 = 525
剩下的空瓶就是：（第一波剩下的空瓶/3 + 第一波剩下的空瓶%3）
（364/3 + 364%3）/3+（364/3 + 364%3 ）%3
122 / 3 + 122 % 3 = 42
这时候可以发现，喝的可乐会越来越多，剩下的空瓶会越来越少；
那么什么时候这个循环结束呢，就是空瓶不足以再换新的可乐的时候，也就是空瓶数量 < 3 的时候；
已知有364瓶可乐，那么设置可乐数量：int sum=364；
瓶子的数量：int ping = sum; //瓶子的数量和可乐数量是相同的
这时进行循环，只要瓶子的数量大于等于3（可以兑换新可乐），那么就执行循环；
while (ping >= 3) { sum += ping / 3; ping = ping / 3+ ping % 3; }
然后输出一共可以喝多少瓶可乐，剩下多少空瓶就可以了
源代码如下：
int sum = 364; int ping = sum; while (ping >= 3) { sum += ping / 3; ping =
ping / 3+ ping % 3; } Console.WriteLine("一共可以喝多少瓶可乐：" + sum + "\n还剩多少瓶：" +
ping);
运行结果如下：
这道题目主要是研究数学算法，逻辑还是需要好好学习的呀！
最后给题目添加了一个小功能，可以输入可乐的数量和兑换方式：
Console.WriteLine("现在有几瓶可乐？"); int sum = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("几瓶可乐能整一瓶？"); int value =
Convert.ToInt32(Console.ReadLine()); int ping = sum; while (ping >= value) {
sum += ping / value; ping = ping / value + ping % value; }
Console.WriteLine("一共可以喝多少瓶可乐：" + sum + "\n还剩多少瓶：" + ping);
运行结果是这样子的：
制作不易，来个一键三连叭！！
