四个女人色诱狱警后越狱后过河的问题 !

Question

2: 话说有四个女人越狱,都很有姿色,把一个看门狱警骗上床,然后穿着睡衣逃跑,跑到一条河边,有一座桥，夜间有一火把，每次最多过两个，必需带火把,过桥速度不一样 
no.1 1min 
no.2 2min 
no3 5min 
no.4 10min 
两个人过用最慢一个的速度，火把不能扔 
如何在17min内四个女人都过桥 之前的题目不够吸引人?都没有人看,我试试有女人的标题有多少人看.

Answer 1

1、NO1 + NO2 过
2、NO1 回
3、NO3、NO4过
4、NO2回
5、NO1、NO2过

Answer 2

题目是老,但做法各不相同:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;namespace WomenBridge
{
    class Program
    {
        static void Main(string[] args)
        {
            //List<Woman> left = new List<Woman>() { new Woman(1, 1), new Woman(2, 2), new Woman(3, 5), new Woman(4, 10) };
            List<Woman> left = new List<Woman>() { new Woman(1, 1), new Woman(2, 2), new Woman(3, 5), new Woman(4, 10),
                new Woman(5,3),new Woman (6,8),new Woman(7,15),new Woman(8,20) };
            FuckingBridge fb=new FuckingBridge ();
            fb.Compute(left);
            Console.ReadLine();
        }
        
        
        #region Woman
        public class FuckingBridge
        {
            public void Compute(List<Woman> left)
            {
                left.Sort();
                List<Woman> right = new List<Woman>();                if (left.Count <= 0)
                {
                    Console.WriteLine("没有人要过河!");
                }
                else if (left.Count == 1)
                {
                    Console.WriteLine("{0}自己过河,耗时{1}", left[0].Id, left[0].CrossTime);
                }
                else
                {
                    int sum = 0;                    while (left.Count > 0)
                    {
                        sum += firstClassStep(left, right);
                        sum += SecondClassStep(left, right);
                    }
                    Console.WriteLine("总耗时:" + sum);
                }
            }
            private int firstClassStep(List<Woman> left, List<Woman> right)
            {
                int time = 0;
                if (left.Count > 0)
                {
                    Woman w0 = left[0];
                    Woman w1 = left[1];                    if (left.Count == 2)
                    {
                        //过去就不用回来.
                        right.Insert(0, w0);
                        right.Insert(1, w1);
                        left.Clear();
                        time = w1.CrossTime;
                        Console.WriteLine("{0}和{1}一起过河,无返回.耗时:{2}", w0, w1, time);
                    }
                    else
                    {
                        //只有w1过去而已.
                        right.Insert(0, w1);
                        left.RemoveAt(1);
                        time = w0.CrossTime + w1.CrossTime;
                        Console.WriteLine("{0}和{1}一起过河,{0}返回.耗时:{2}", w0, w1, time);
                    }
                }
                return time;
            }
            private int SecondClassStep(List<Woman> left, List<Woman> right)
            {
                int time = 0;
                int size = left.Count;
                if (size > 0)
                {
                    Woman w0 = left[size - 2];
                    Woman w1 = left[size - 1];
                    left.RemoveAt(size - 1);
                    left.RemoveAt(size - 2);
                    right.Insert(1, w0);
                    right.Insert(2, w1);
                    if (left.Count == 0)
                    {
                        time = w1.CrossTime;
                        Console.WriteLine("{0}和{1}一起过河,无返回.耗时:{2}", w0, w1, time);
                    }
                    else
                    {
                        Woman w2 = right[0];
                        right.RemoveAt(0);
                        left.Insert(1, w2);
                        time = w1.CrossTime + w2.CrossTime;
                        Console.WriteLine("{0}和{1}一起过河,{2}返回.耗时:{3}", w0, w1, w2, time);
                    }
                }
                return time;
            }
        }
        public class Woman : IComparable<Woman>
        {
            public Woman(int id, int time)
            {
                this.Id = id;
                this.CrossTime = time;
            }
            public int Id { get; set; }
            public int CrossTime { get; set; }
            public override string ToString()
            {
                return string.Format("(Id:{0},Time:{1})", Id.ToString().PadLeft(3, ' '), CrossTime.ToString().PadLeft(4, ' '));
            }
            #region IComparable<Woman> Members            public int CompareTo(Woman other)
            {
                if (other == null) return 1;
                if (this.CrossTime > other.CrossTime) return 1;
                if (this.CrossTime < other.CrossTime) return -1;
                return 0;
            }            #endregion
        }
        #endregion
    }
}

Answer 3

过法如下：
第一个来回：
  1：no.1和no.2 过桥，no.2留下。 耗时为：2秒
  2：no.1把火把带回来，耗时为：1秒
第二个来回：
  1：no.3和no.4过桥，no.3和no.4都留下来。耗时为：10秒
  2：no.2把火把带回来，耗时为：2秒
第三个来回：
  1：no.1和no.2 过桥。 耗时为：2秒
 经过三次全部都过桥了。总共耗时：2+1+10+2+2=17.
明白了吗？

Answer 4

d[0]+d[1]+d[2]+...d[1000]-1-2-3-...-1000 =?

Answer 5

static void Main(string[] args)
 2        {
 3            int[] list = new int[1001];
 4
 5            for (int i = 1; i < 1001; i++)
 6            {
 7                list[i - 1] = i;
 8            }
 9
10            Random random = new Random();
11
12            list[1000] = random.Next(1, 1000);
13
14            int sum1 = 0;
15
16            int sum2 = 0;
17
18            foreach (int i in list)
19            {
20
21                sum1 = sum1 + i;
22            }
23            for (int i = 1; i < 1001; i++)
24            {
25
26                sum2 = sum2 + i;
27            }
28
29
30            Console.WriteLine("重复的数字是:"+(sum1 - sum2).ToString());
31            Console.Read();
32        }