[sicily]1052. Candy Sharing Game

最新推荐文章于 2019-06-26 23:26:53 发布

大笨猪耶

最新推荐文章于 2019-06-26 23:26:53 发布

阅读量346

点赞数

CC 4.0 BY-SA版权

分类专栏：数据结构与算法分析文章标签： sicily

本文链接：https://blog.youkuaiyun.com/luohuang0423/article/details/49556107

数据结构与算法分析专栏收录该内容

58 篇文章

订阅专栏

本文介绍了一个基于学生间传递糖果的游戏模拟问题，通过环状数组记录每位学生的糖果数量，并利用循环来模拟游戏进程，最终确定游戏结束时各学生拥有的糖果数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1052. Candy Sharing Game

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

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


Notes:
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

简单的游戏模拟题，利用数组就可以轻松解决。代码如下：

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;

int main()
{
    int t;
    
    while(cin>>t && t>0)
    {
        int circle[t];
        int count=0,flag=1;
        for(int i=0; i<t; i++)
            cin>>circle[i];
        while(flag==1)
        {
            flag = 0;
            if(circle[0] != circle[t-1])//先判断圈中是否存在不同数，如果有设 flag = 1;
                flag = 1;
            for(int i=0; i<t-1; i++)
            {
                if(circle[i] != circle[i+1])
                    flag = 1;
            }
            if(flag==1)   // 有不相同的数，继续将自己的一半分给右边的人
            {
                count++;
                int tmp = circle[t-1];
                for(int i=t-1; i>0; i--)
                {
                    circle[i] = (circle[i]+circle[i-1])/2;
                }
                circle[0] = (circle[0]+tmp)/2;
                for(int i=0; i<t; i++) // 如果奇数，从老师那边获取一个，凑成偶数个
                {         
                    if(circle[i]%2 == 1)
                        circle[i]++;
                }
            }
        }   
        cout<<count<<" "<<circle[0]<<endl;
    } 
    //system("pause");
    return 0;   
}