Math Puzzles

数学题

下面这个链接里有71道数学题：

http://www.math.ucsb.edu/~cooper/puz1.html

谁谁有兴趣就做做。我只做过第1道和第9道。

第1道：

view plainprint?

1.  /** 
2.  * MATH PUZZLE 01 
3.  * 
4.  * An ant starts at one end of a rubber band and walks along it at a speed 
5.  * of 1 inch per second. The rubber band is 10 inches long, and is being 
6.  * stretched uniformly at a rate of 20 inches per second. Does the ant ever 
7.  * reach the other end of the rubber band?. Give reasons. 
8.  */  
9. public void testMathPuzzle01() {  
10.     long t = 0;  
11.     final long MINUTE = 60, HOUR = MINUTE * 60, DAY = HOUR * 24, YEAR = DAY * 365;  
12.     double rubber = 10;  
13.     double walked = 0;  
14.     while (walked < rubber) {  
15.         // After 1 second  
16.         t++;  
17.         rubber = rubber + 20;  
18.         walked = walked * rubber / (rubber - 20) + 1;  
19.         if (t % YEAR == 0) {  
20.             System.out.println("" + t / YEAR + " years has been flying by");  
21.         }  
22.     }  
23.     System.out.println("Time used:" + t / YEAR + " years, " + t % YEAR  
24.             / DAY + " days, " + t % DAY / HOUR + " hours," + t % HOUR  
25.             / MINUTE + " minutes," + t % MINUTE + " seconds");  
26. }  

    /**
    * MATH PUZZLE 01
    *
    * An ant starts at one end of a rubber band and walks along it at a speed
    * of 1 inch per second. The rubber band is 10 inches long, and is being
    * stretched uniformly at a rate of 20 inches per second. Does the ant ever
    * reach the other end of the rubber band?. Give reasons.
    */
   public void testMathPuzzle01() {
       long t = 0;
       final long MINUTE = 60, HOUR = MINUTE * 60, DAY = HOUR * 24, YEAR = DAY * 365;
       double rubber = 10;
       double walked = 0;
       while (walked < rubber) {
           // After 1 second
           t++;
           rubber = rubber + 20;
           walked = walked * rubber / (rubber - 20) + 1;
           if (t % YEAR == 0) {
               System.out.println("" + t / YEAR + " years has been flying by");
           }
       }
       System.out.println("Time used:" + t / YEAR + " years, " + t % YEAR
               / DAY + " days, " + t % DAY / HOUR + " hours," + t % HOUR
               / MINUTE + " minutes," + t % MINUTE + " seconds");
   }

蚂蚁这题很有意思。应该是一个微积分的题。上面的解答算一个模拟解吧。假设先拉橡皮筋，蚂蚁后走：

walked = walked * rubber / (rubber - 20) + 1;

算出蚂蚁需要大约16年才能爬到橡皮筋另一头。或者蚂蚁先走，再拉橡皮筋：

walked = (walked + 1) * rubber / (rubber - 20) ;

可以算出蚂蚁需要2年多一点可以爬到另一头。早一步就节省12年。这道题也印证了抢占先机的重要性。

还有就是第9道：

view plainprint?

1. /** 
2.   * MATH PUZZLE 09 
3.   * 
4.   * In a TV show, a prize is hidden behind one of 3 closed doors. The 
5.   * contestant tries to guess where the prize is. After the contestant 
6.   * chooses a door, the host of the show (who knows where the prize is) opens 
7.   * one of the 2 remaining doors, to reveal that the prize is not behind that 
8.   * door. The host then gives the contestant the opportunity to change her 
9.   * guess. Should she? 
10.   */  
11.  public void testMathPuzzle09() {  
12.      final int SIZE = 100000;  
13.      boolean[] doors = new boolean[3];  
14.      Random random = new Random();  
15.      int insist = 0;  
16.      int reselect = 0;  
17.      for (int i = 0; i < SIZE; i++) {  
18.          // Init doors  
19.          doors[0] = doors[1] = doors[2] = false;  
20.          int index = random.nextInt(3);  
21.          doors[index] = true;  
22.          int guess = random.nextInt(3);  
23.          // does not change her guess  
24.          if (doors[guess] == true) {  
25.              insist++;  
26.          }  
27.          // change her guess  
28.          if (doors[guess] == false) {  
29.              reselect++;  
30.          }  
31.      }  
32.      System.out.println("Not change will get:" + insist);  
33.      System.out.println("Change will get:" + reselect);  
34.  }   

  /**
    * MATH PUZZLE 09
    *
    * In a TV show, a prize is hidden behind one of 3 closed doors. The
    * contestant tries to guess where the prize is. After the contestant
    * chooses a door, the host of the show (who knows where the prize is) opens
    * one of the 2 remaining doors, to reveal that the prize is not behind that
    * door. The host then gives the contestant the opportunity to change her
    * guess. Should she?
    */
   public void testMathPuzzle09() {
       final int SIZE = 100000;
       boolean[] doors = new boolean[3];
       Random random = new Random();
       int insist = 0;
       int reselect = 0;
       for (int i = 0; i < SIZE; i++) {
           // Init doors
           doors[0] = doors[1] = doors[2] = false;
           int index = random.nextInt(3);
           doors[index] = true;
           int guess = random.nextInt(3);
           // does not change her guess
           if (doors[guess] == true) {
               insist++;
           }
           // change her guess
           if (doors[guess] == false) {
               reselect++;
           }
       }
       System.out.println("Not change will get:" + insist);
       System.out.println("Change will get:" + reselect);
   } 

实质上是大家常讲的三个门的题。门后面可能藏别的东西。