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\n",sum++,a[n]);
}
return 0;
}

模拟:HDU1034-Candy Sharing Game的更多相关文章

  1. HDU-1034 Candy Sharing Game 模拟问题(水题)

    题目链接:https://cn.vjudge.net/problem/HDU-1034 水题 代码 #include <cstdio> #include <algorithm> ...

  2. HDU1034 Candy Sharing Game

    Problem Description A number of students sit in a circle facing their teacher in the center. Each st ...

  3. Candy Sharing Game(模拟搜索)

    Candy Sharing Game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Other ...

  4. POJ - 1666 Candy Sharing Game

    这道题只要英语单词都认得,阅读没有问题,就做得出来. POJ - 1666 Candy Sharing Game Time Limit: 1000MS Memory Limit: 10000KB 64 ...

  5. hdu 1034 Candy Sharing Game

    Candy Sharing Game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Other ...

  6. HDU 1034 Candy Sharing Game (模拟)

    题目链接 Problem Description A number of students sit in a circle facing their teacher in the center. Ea ...

  7. 九度OJ 1145:Candy Sharing Game(分享蜡烛游戏) (模拟)

    时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:248 解决:194 题目描述: A number of students sit in a circle facing their teac ...

  8. M - Candy Sharing Game

    Description A number of students sit in a circle facing their teacher in the center. Each student in ...

  9. Candy Sharing Game(hdoj1034)

    Problem Description A number of students sit in a circle facing their teacher in the center. Each st ...

随机推荐

  1. 开发原则&设计模式

    1.关于软件开发中的开发原则和设计模式: 1.1.开发原则 1.1.1.什么是开发原则? 开发原则就是开发的依据,只要依照这些原则进行开发,将来开发的软件具有很强的扩展力,很低的耦合度. 开发原则不属 ...

  2. html5标签的兼容性处理

    HTML5的语义化标签以及属性 1.可以让开发者非常方便地实现清晰的web页面布局,加上CSS3的效果渲染,快速建立丰富灵活的web页面显得非常简单 2.使用他们能让代码语义化更直观,而且更方便SEO ...

  3. 树莓派-(一)开箱到点亮一些坑(无屏、无wlan、无直连键鼠)

    0x00.前期准备: 材料: 树莓派3b+ 板子 * 1,适配电源 * 1,网线 * 2,sd卡16G * 1,读卡器 * 1 安装时注意,3b+三个散热片贴好.小风扇接线要接对 工具: 0x01. ...

  4. WPF创建SignalR服务端(转)

    在网上看到了一个帖子,比较详细,博主写的很好. 地址:http://blog.csdn.net/lordwish/article/details/51786200

  5. 使用腾讯IP分享计划网站中的纯JS省市区三级联动

    JS地址:http://ip.qq.com/js/geo.js 实例如下: <!DOCTYPE html> <html> <head> <title>省 ...

  6. echo -e的用法

    root@bt:~# echo -e "HEAD /HTTP/1.0\n\n"HEAD /HTTP/1.0 root@bt:~# echo -e "HEAD /HTTP/ ...

  7. Mantis-1.3.3 (Ubuntu 16.04)

    平台: Ubuntu 类型: 虚拟机镜像 软件包: mantis-1.3.3 bug tracking commercial devops mantis open-source project man ...

  8. 使用CreateProcess函数运行其他程序

    为了便于控制通过脚本运行的程序,可以使用win32process模块中的CreateProcess()函数创建一个运行相应程序的进程.其函数原型如下.CreateProcess(appName, co ...

  9. pat甲级1012

    1012 The Best Rank (25)(25 分) To evaluate the performance of our first year CS majored students, we ...

  10. c++ vector & 二维数组 & MessageBox

    vector: https://www.cnblogs.com/mr-wid/archive/2013/01/22/2871105.html c++ 二维数组: int **p; p = new in ...