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 64bit IO Format: %I64d & %I64u
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
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
Source
Greater New York 2003
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h> int n, stu[], give[]; int main()
{
while(scanf("%d", &n)) {
if(!n) break;
for(int i = ; i < n; i++) {
scanf("%d", &stu[i]);
}
int time = ;
while(true) {
bool flag = ;
for(int i = ; i < n; i++) {
if(i > && stu[i] != stu[]) {
flag = ;
break;
}
}
if(!flag) break; time++; for(int i = ; i < n; i++) {
give[i] = stu[i]/;
} for(int i = ; i < n; i++) {
stu[i] = give[i] + give[(i+)%n];
if(stu[i]%) {
stu[i]++;
}
} } printf("%d %d\n", time, stu[]); } return ;
}
POJ - 1666 Candy Sharing Game的更多相关文章
- hdu 1034 Candy Sharing Game
Candy Sharing Game Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- Candy Sharing Game(模拟搜索)
Candy Sharing Game Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- 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 ...
- HDU1034 Candy Sharing Game
Problem Description A number of students sit in a circle facing their teacher in the center. Each st ...
- 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 ...
- sicily 1052. Candy Sharing Game
Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description A number of students sit in a circle ...
- 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 ...
- POJ 1666
#include<iostream> using namespace std; int main() { int num_stu; int i; ; do{ time=; cin>& ...
随机推荐
- Git初使用
今天开始初次使用Git,Git作为一个使用广泛的分布式版本控制系统,我们有必要熟悉掌握. 这次主要是实现将本地上的“Hello World”的完整的项目文件提交到github上新建的代码库,主要过程如 ...
- Ansible (一)
epel rpm -ivh http://mirrors.ustc.edu.cn/fedora/epel/6/x86_64/epel-release-6-8.noarch.rpm yum -y ins ...
- psutil 是因为该包能提升 memory_profiler 的性能
python 性能分析入门指南 一点号数据玩家昨天 限时干货下载:添加微信公众号"数据玩家「fbigdata」" 回复[7]免费获取[完整数据分析资料!(包括SPSS.SAS.SQ ...
- 【转】NumPy-快速处理数据
2.0 简介 标准安装的Python中用列表(list)保存一组值,可以用来当作数组使用,不过由于列表的元素可以是任何对象,因此列表中所保存的是对象的指针(为了保存各种类型的对象,只能牺牲空间).这样 ...
- 使用eclipse开发servlet
package cn.itcast; import java.io.IOException; import javax.servlet.GenericServlet; import javax.ser ...
- JS控制打印指定div
<html><head><script language="javascript">function printdiv(printpage){v ...
- Memcached 笔记与总结(9)Memcached 与 Session
一.Memcached 存储 Session 由于 Memcached 是分布式的内存对象缓存系统,因此可以用来实现 Session 同步:把 Web 服务器中的内存组合起来,成为一个“内存池”,不管 ...
- php高并发状态下文件的读写
php高并发状态下文件的读写 背景 1.对于PV不高或者说并发数不是很大的应用,不用考虑这些,一般的文件操作方法完全没有问题 2.如果并发高,在我们对文件进行读写操作时,很有可能多个进程对进一文件 ...
- DotNetBar中ListViewEx控件的使用
最近一直在学习DotNetBar,今天遇到的问题是ListView的使用问题,其实没有特别难的,只是写在这里给自己留个记录. 首先,在Form中添加一个ListViewEx控件, 初始化中写代码如下: ...
- lr并发量和迭代的区别
1.并发量 并发量也就是同时运行的量.比如100个用户同时登录,那么并发量就是100.当然这100个用户可以进行参数化,也可以采用设置虚拟用户数(vuser). 2.迭代 迭代就是单个用户运行的次数. ...