1034

Candy Sharing Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4629    Accepted Submission(s): 2832

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
   
   
   
   

    就是跟着题意模拟就是了，我这里使用变量a记录上一个小孩给的糖果个数。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

#define MAXN 100010

using namespace std;

int n;
int kkke[MAXN];

bool not_equal()
{
	for(int i=1;i<n;i++)
	    if(kkke[i]!=kkke[0])
	        return true;
	return false;
}

int main()
{
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	while(scanf("%d",&n),n)
	{
		int a,b,cnt=0;
		for(int i=0;i<n;i++)scanf("%d",&kkke[i]);
		while(not_equal())
		{
			cnt++;
			kkke[0]>>=1;
			a=kkke[0];
			for(int i=1;i<n;i++)
			{
				b=kkke[i]>>1;
				kkke[i]=a+b;
				if(kkke[i]&1)kkke[i]++;
				a=b;
			}
			kkke[0]+=a;
			if(kkke[0]&1)kkke[0]++;
		}
		printf("%d %d\n",cnt,kkke[0]);
	}
	return 0;
}