解题心得：
1、直接模拟每一次分一半就行了，模拟过程，记录轮数，但是也看到有些大神使用的是链表，估计链表才是真的做法吧。
题目：
Candy Sharing Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6525 Accepted Submission(s): 3962
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
Hint
The game ends in a finite number of steps because:
1. The maximum candy count can never increase.
2. The minimum candy count can never decrease.
3. No one with more than the minimum amount will ever decrease to the minimum.
4. If the maximum and minimum candy count are not the same, at least one student with the minimum amount must have their count increase.
#include<bits/stdc++.h>
using namespace std;
int a[10000],b[10000];//b用于记录每次分一半的数目；
int main()
{
int n;
bool flag;//记录是否已经分配平衡；
while(~scanf("%d",&n))
{
if(n == 0)
break;
int sum = 0;
flag = false;
//初始化输入
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(a[i]%2)
a[i]+=1;
}
while(!flag)
{
sum ++;
for(int i=1;i<=n;i++)
{
b[i] = a[i]/2;
a[i]/=2;
}
//模拟过程
for(int i=2;i<=n;i++)
a[i] = a[i] + b[i-1];
a[1] = a[1] + b[n];
for(int i=1;i<=n;i++)
if(a[i]%2)
a[i]++;//是单数的加一
int k;
for(k=1;k<n;k++)
if(a[k] != a[k+1])
break;
if(k == n)
flag = true;
}
printf("%d %d
",sum++,a[n]);
}
return 0;
}