POJ 1666 Candy Sharing Game（模拟）

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本文介绍了一种名为Candy Sharing Game的游戏算法实现。该游戏中，孩子们围坐一圈，在老师的哨声下将手中的糖果平均传递给右侧的孩子，并通过老师的补充使糖果数保持偶数，直至每个孩子拥有相同数量的糖果。文章提供了具体的解题思路及C++实现代码。

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Candy Sharing Game

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

Sample Output

15 14
17 22
4 8

题目大意：有n个小孩围着老师坐成一圈，每个小孩有偶数个糖果，当老师吹哨子的时候，每个小孩都把自己手里糖果的一半给右手边的孩子，如果出现奇数，则老师会补一个糖果，凑成偶数。重复此过程直到每个孩子拥有的糖果数目相同，求吹哨子的次数和最后每个孩子所拥有的糖果数。

解题思路：开一个数组存每个孩子的糖果数量，从最后一个孩子开始处理，吹一次哨子后，

candy[i] += candy[i - 1] / 2 + candy[i] / 2 (i > 0)

candy[0] += candy[n - 1] / 2 + candy[0] / 2 (i = 0)

然后判断candy[i]是否为偶数，若不为偶数，则加一。

代码如下：

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

using namespace std;
int candy[1000];

int main()
{
	int n,i,j,k;
	int add,del;
	int max,min,maxpos,minpos;
	while(scanf("%d",&n) != EOF && n){
		memset(candy,0,sizeof(candy));
		for(i = 0;i < n;i++){
			scanf("%d",&candy[i]);
		}
		int cnt = 0;
		
		while(true){
			cnt++;
			add = candy[n - 1] / 2;
			for(i = n - 1;i > 0;i--){
				candy[i] += candy[i - 1] / 2 - candy[i] / 2;
				if(candy[i] % 2 == 1)
					candy[i] += 1; 
			}
			candy[0] += add - candy[0] / 2;
			if(candy[0] % 2 == 1)
				candy[0] += 1;
			
			for(i = 1;i < n;i++){
				if(candy[i] != candy[0])
					break;
			}
			//如果i == n,说明candy数组中所有元素均相等，退出循环 
			if(i == n)
				break;
		}
		printf("%d %d\n",cnt,candy[0]);
	}
	return 0;
 }