穷举 迭代

//1、一张150元购物卡，三类洗化用品，洗发水15元，香皂2元，牙刷5元
 //求刚好花完150元，有多少种买法，每种买法各买几样？
 //洗发水 x    香皂 z   牙刷 y 
 int count = 0;
 int bian = 0;
 for (int x = 0; x <= 10; x++)
 {
       for (int y = 0; y <= 30; y++)
       {
             for (int z = 0; z <= 75; z++)
             {
                 bian++;
                 if (15 * x + 5 * y + 2 * z == 150)
                 {
                      count++;
                      Console.WriteLine("第" + count + "种买法洗发水" + x + "瓶，香皂" + z + "块，牙刷" + y + "支");
                 }
             }
       }
 }
 Console.WriteLine("一共循环了" + bian + "遍");
 Console.WriteLine("一共有" + count + "种买法");
 Console.ReadLine();

 //公鸡二文钱一只，母鸡一文钱一只，小鸡半文钱一只，现在有100文钱，要买一百只鸡，
 //正好花完100文钱,共有多少种买法，每种各买几只？
 int count = 0;
 int bian = 0;
 for (int gong = 0; gong <= 50; gong++)
 {
     for (int mu = 0; mu <= 100; mu++)
     {
          for (int xiao = 0; xiao <= 200; xiao++)
          {
                bian++;
                if (gong + mu + xiao == 100 && gong * 2 + mu * 1 + xiao * 0.5 == 100)
                {
                     count++;
                     Console.WriteLine("第" + count + "种买法公鸡" + gong + "只，母鸡" + mu + "只，小鸡" + xiao + "只");
                }
          }
     }
 }
 Console.WriteLine("一共循环了" + bian + "遍");
 Console.WriteLine("共有" + count + "种买法");
 Console.ReadLine();

//for循环变形得到while循环
//初始条件拿到前面，循环的状态改变放到循环体的最后一句
//for变成while，将原先的分号去掉，只留下循环条件
//数字1~10，累加求和
int sum = 0;
int i = 1;
while (i <= 10)
{
     sum += i;
     i++;
}
Console.WriteLine(sum);

//公鸡二文钱一只，母鸡一文钱一只，小鸡半文钱一只，现在有100文钱，要买一百只鸡，
//正好花完100文钱,共有多少种买法，每种各买几只？ 用while循环
int count = 0;
int bian = 0;
int gong = 0;
while (gong <= 50)
{
    int mu = 0;
    while (mu <= 100)
    {
         int xiao = 0;
         while (xiao <= 200)
         {
             bian++;
             if (gong + mu + xiao == 100 && gong * 2 + mu * 1 + xiao * 0.5 == 100)
             {
                  count++;
                  Console.WriteLine("第" + count + "种买法公鸡" + gong + "只，母鸡" + mu + "只，小鸡" + xiao + "只");
             }
             xiao++;
         }
          mu++;
    }
    gong++;
}
Console.WriteLine("共有" + count + "种买法");
Console.WriteLine("共循环了" + bian + "遍");
Console.ReadLine()；

//迭代
//纸张厚度0.07毫米
//折叠多少次可以超过珠峰高度8848米
double a = 0;
double i = 0.07;
while (i <= 8848000)
{
    a++;
    i *= 2;
}
Console.WriteLine(a);

//大马驼2石粮食，中等马驼1石粮食，两头小马驼1石粮食，要用100匹马，驼100石粮食，该如何分配？
int count = 0;
int bian = 0;
for (int dm = 0; dm <= 50; dm++)
{
     for (int zm = 0; zm <= 100; zm++)
     {
          for (int xm = 0; xm <= 200; xm++)
          {
               bian++;
               if (dm + zm + xm == 100 && dm * 2 + zm * 1 + xm * 0.5 == 100)
               {
                     count++;
                     Console.WriteLine("第" + count + "种方法大马" + dm + "匹，中等马" + zm + "匹，小马" + xm + "匹");
               }
          }
     }
}
Console.WriteLine("一共循环了" + bian + "遍");
Console.WriteLine("一共有" + count + "种方法");
Console.ReadLine();

//有1分钱，2分钱，5分钱的硬币，要组合出来2角钱，有几种组合方式，分别各多少个？
int count = 0;
int bian = 0;
for (int yf = 0; yf <= 20; yf++)
{
     for (int ef = 0; ef <= 10; ef++)
     {
          for (int wf = 0; wf <= 4; wf++)
          {
               bian++;
               if (yf * 1 + ef * 2 + wf * 5 == 20)
               {
                   count++;
                   Console.WriteLine("第" + count + "种组合方式1分硬币" + yf + "个，2分硬币" + ef + "个，5分硬币" + wf + "个");
               }
          }
     }
}
Console.WriteLine("共有" + count + "种组合方式");
Console.WriteLine("共循环了" + bian + "遍");
Console.ReadLine();

//输入一个100以内的整数，求从1加到这个数的和，如果输入的数不在0到100之间，提示请重新输入
for (; ; )
{
    Console.Write("请输入一个100以内整数：");
    int a = int.Parse(Console.ReadLine());
    int sum = 0;
    if (a >= 0 && a <= 100)
    {
        for (int i = 0; i <= a; i++)
        {
             sum += i;
        }
        Console.WriteLine(sum);
        break;
    }
    else
    {
        Console.WriteLine("您的输入有误，请重新输入");
    }
}
Console.ReadLine();