Candy Sharing Game

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.

题意：n个学生围成一个圈坐在一起，老师分别给每个人一定量的糖果，然后进行分糖。

要求：1.如果存在某一个学生的糖果数与其他人的不同，则所有人同时将自己糖果数量的一半给他的左边或者右边。

            2.当某个学生的糖果数量为单数时，可以从老师处获得一个糖果。

最终使所有的学生得到的糖果数相同，问：需要分糖果的次数以及最终每个学生得到的糖果数。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[1000];
int n;
///暴力求解即可
int f(){
    int i;
    for(i=0;i<n-1;i++){
        if(a[i]!=a[i+1])return 0;
    }
    return 1;
}

int main(){

while(cin>>n&&n!=0){
      int cnt=0;
      memset(a,0,sizeof(a));
      int i,j;

      for(i=0;i<n;i++){
            cin>>a[i];
      }
      while(f()==0){
            cnt++;
            int t = a[0];
            for(i=0;i<n-1;i++){
                  a[i]=a[i]/2+a[i+1]/2;
                  if(a[i]%2==1)a[i]++;
            }
            a[n-1] = t/2+a[n-1]/2;

            if(a[n-1]%2==1)a[n-1]++;
      }
      cout<<cnt<<" "<<a[0]<<endl;
}

return 0;
}