杭电ACM OJ 1034 Candy Sharing Game 水水更健康（只是题目意思有点难理解）

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

  
  

  
  
   翻译：
  
比如

36

2

2

2

2

2


代表6个小孩围成一圈，分别持有不同数量的糖果。36 2 2 。。。

现在所有小孩要同时顺时针分一半的糖给下一个小孩。

比如36就要给18个给2

2就要给1给下一个2

。。

最后一个2就要给1给36（其实是18了已经）

如果给完后发现，某个小孩是奇数个糖果，那么老师就会给他一个糖果，保证每个人手上的糖果数量都是偶数的。




问多少次循环以后所有人的糖果数量是相等的。这个相等的数量又是多少。




思路：没有思路，就这么做。





       
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]直接变为一半并记录
                a[i] /= 2;
                half[i] = a[i];
            }

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

            //先赋值第一个
            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，一定要刷到一题不水的题把它做出来后才不刷了。