杭電ACM OJ 1034 Candy Sharing Game 水水更健康（隻是題目意思有點難了解）

Candy Sharing Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7573    Accepted Submission(s): 4531

Problem Description A number of students sit in a circle facing their teacher in the center. Each student initially has an even number of pieces of candy. When the teacher blows a whistle, each student simultaneously gives half of his or her candy to the neighbor on the right. Any student, who ends up with an odd number of pieces of candy, is given another piece by the teacher. The game ends when all students have the same number of pieces of candy. 

Write a program which determines the number of times the teacher blows the whistle and the final number of pieces of candy for each student from the amount of candy each child starts with.

Input The input may describe more than one game. For each game, the input begins with the number N of students, followed by N (even) candy counts for the children counter-clockwise around the circle. The input ends with a student count of 0. Each input number is on a line by itself.

Output For each game, output the number of rounds of the game followed by the amount of candy each child ends up with, both on one line.

Sample Input

6
36
2
2
2
2
2
11
22
20
18
16
14
12
10
8
6
4
2
4
2
4
6
8
0
        

Sample Output

15 14
17 22
4 8

  
  

  
  
   翻譯：
  
比如

36

2

2

2

2

2


代表6個小孩圍成一圈，分别持有不同數量的糖果。36 2 2 。。。

現在所有小孩要同時順時針分一半的糖給下一個小孩。

比如36就要給18個給2

2就要給1給下一個2

。。

最後一個2就要給1給36（其實是18了已經）

如果給完後發現，某個小孩是奇數個糖果，那麼老師就會給他一個糖果，保證每個人手上的糖果數量都是偶數的。




問多少次循環以後所有人的糖果數量是相等的。這個相等的數量又是多少。




思路：沒有思路，就這麼做。





       
package ACM1000_1099;

public class CandySharingGame1034 {
    // 36 2 2 2 2 2
    void calculate() {
        int[] a = {2, 4, 6, 8};
//        int[] a = {36, 2, 2, 2, 2, 2};
        int len = a.length;

        int[] half = new int[len];

        int times = 0;

        while (true) {
            times ++;

            //取得一半
            for (int i = 0; i < len; i ++) {
                //a[i]直接變為一半并記錄
                a[i] /= 2;
                half[i] = a[i];
            }

            //是否全部相等
            boolean flag = true;

            //先指派第一個
            a[0] += half[len - 1];
            if (a[0] % 2 != 0) {
                a[0] += 1;
            }

            for (int i = 1; i < len; i ++) {
                a[i] += half[i - 1];

                if (a[i] % 2 != 0) {
                    a[i] += 1;
                }

                if (a[i] != a[i - 1]) {
                    flag = false;
                }
            }

            if (flag) {
                break;
            }
        }

        System.out.print(a[0] + " ");
        System.out.print(times + "");
    }

    public static void main(final String[] args) throws Exception {
        CandySharingGame1034 c = new CandySharingGame1034();
        c.calculate();
    }
}
      
        
我不信邪了，一定要刷到一題不水的題把它做出來後才不刷了。