hdu 1034 Candy Sharing Game
Candy Sharing Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4448 Accepted Submission(s):
2718
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.
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.
followed by the amount of candy each child ends up with, both on one
line.
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<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<queue>
#include<map>
#include<vector>
#define MAX 100010
#define LL long long
#define INF 0xfffffff
#define fab(x) (x)>0?(x):(-x)
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int n,m,j,i,t;
int s[MAX],s1[MAX];
while(scanf("%d",&n),n)
{
for(i=1;i<=n;i++)
{
scanf("%d",&s[i]);
s1[i]=s[i];
}
int flag=1;
int ans=0;
int ant=0;
while(flag)
{
ans++; s[1]=s1[n]/2+s1[1]/2;
if(s[1]&1)
s[1]+=1; //s1[1]=s[1]; for(i=2;i<=n;i++)
{
s[i]=s1[i-1]/2+s1[i]/2;
if(s[i]&1)
s[i]+=1;
//s1[i]=s[i];
}
for(i=1;i<=n;i++)
s1[i]=s[i];
// for(i=1;i<=n;i++)
// {
// printf("%d# ",s[i]);
// }
// printf("\n"); int k=0;
for(i=1;i<n;i++)
{
if(s[i]==s[i+1])
k++;
else
break;
}
for(i=1;i<=n;i++)
printf("%d ",s[i]);
printf("\n");
if(s[1]==s[n])
k++;
if(k==n)
{
flag=0;
ant=s[1];
}
}
printf("%d %d\n",ans,ant);
}
return 0;
}
hdu 1034 Candy Sharing Game的更多相关文章
- 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 ...
- 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 ...
- Candy Sharing Game(模拟搜索)
Candy Sharing Game Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- Hdu 4465 Candy (快速排列组合+概率)
题目链接: Hdu 4465 Candy 题目描述: 有两个箱子,每个箱子有n颗糖果,抽中第一个箱子的概率为p,抽中另一个箱子的概率为1-p.每次选择一个箱子,有糖果就拿走一颗,没有就换另外一个箱子. ...
- HDU 5291 Candy Distribution DP 差分 前缀和优化
Candy Distribution 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5291 Description WY has n kind of ...
- HDU 4780 Candy Factory
Candy Factory Time Limit: 2000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ...
- hdu 4465 Candy(二次项概率)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4465 参考博客:http://www.cnblogs.com/goagain/archive/2012 ...
- M - Candy Sharing Game
Description A number of students sit in a circle facing their teacher in the center. Each student in ...
- 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
- [模拟]ZOJ3480 Duck Typing
题意:给了一坨...按题目意思输出就好了... 给一组案例 begin class d class c:d class b:c class a:b def d.m def d.n call a.m e ...
- 【转】windows c++获取文件信息——_stat函数的使用
_stat函数的功能 _stat函数用来获取指定路径的文件或者文件夹的信息. 函数声明 int _stat( const char *path, struct _stat *buffer ); 参数: ...
- Tomcat下server.xml中context介绍
conf/Context.xml是Tomcat公用的环境配置;若在server.xml中增加<Context path="/test" docBase="D:\te ...
- 221. Maximal Square
题目: Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 1's and ...
- nigix以及相关
nginx+php的配置 php与nginx整合 http://www.thinkphp.cn/topic/13082.html [入门篇]Nginx + FastCGI 程序(C/C++) 搭建高性 ...
- 怎样开发Chrome浏览器的插件
http://jingyan.baidu.com/article/b907e627fb90fd46e7891c3c.html Chrome 浏览器作为基于Webkit的新一代浏览器.Chrome自从正 ...
- MySQL数据库乱码 - Linux下乱码问题一
乱码问题是很让人抓狂的问题,下面我将记录一下linux下mysql乱码问题的解决方法. mysql在linux下乱码问题 一.操作 mysql默认字符集是latin1,但是我们大部分程序使用的字符集是 ...
- Android 签名(2)签名知识要点
要点 1) 所有的应用程序都必须有数字证书,Android系统不会安装一个没有数字证书的应用程序 2) Android程序包使用的数字证书可以是自签名的,不需要一个权威的数字证书机构签名认证 3) 如 ...
- git clone 出错SSL certificate problem, verify that the CA cert is OK.
先调用这个 export GIT_SSL_NO_VERIFY=true 之后再执行git clone
- mysql运算符的优先级
Operator precedences are shown in the following list, from highest precedence to the lowest. Operato ...