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
- WinForm 快捷键设置(转载)
1.Alt+*(按钮快捷键) 按钮快捷键也为最常用快捷键,其设置也故为简单.在大家给button.label.menuStrip等其他控件的Text属性指定名称时,在其后面加上‘&’然后在加上 ...
- 基于 Node.js 平台,快速、开放、极简的 web 开发框架。
资料地址:http://www.expressjs.com.cn/ Express 基于 Node.js 平台,快速.开放.极简的 web 开发框架. $ npm install express -- ...
- 关于webdriver中弹出框的定位
对于webdriver中弹出框的定位,需要引入这个包: import org.openqa.selenium.Alert; 具体语句为: Alert alert= driver.switchTo(). ...
- Sumsets
题目连接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=91209#problem/A 题意:给定一个整数几何S,找出一个最大的d,使得a ...
- ThreadLocal解析
ThreadLocal 如果定义了一个单实例的java bean,它有若干属性,但是有一个属性不是线程安全的,比如说HashMap.并且碰巧你并不需要在不同的线程中共享这个属性,也就是说这个属性不存在 ...
- Hive_进阶
回顾: hive 优点 1. 类sql语句靠近关系型数据库,可自定义函数,增加了扩展性,易于开发,减少mapreduce学习成本 2. hive转换sql语句为mapreduce程序以mapreduc ...
- cursor 手型样式
cursor:hand 与 cursor:pointer 的效果是一样,都像手形光标.但用FireFox浏览时才注意到使用cursor:hand在FireFox里并被支持.cursor:hand :I ...
- Lua修改文件名
local lfs = require "lfs"local curPath = "..\\new\\"local totalFile = 0 local s ...
- HTTP 战役 与 历史
导火线1992年,有一家公司Nombas 开发了一种叫C--的嵌入式脚本语言,后来觉得名字比较晦气,最终改名为scriptEase.而这种可以嵌入网页中的脚本的理念,成为日后移动互联网蓬勃发展的一块重 ...
- ngrok访问外网
1. 外网映射工具介绍 windows用户: 1,下载windows版本的客户端,解压到你喜欢的目录2,在命令行下进入到path/to/windows_386/下3,执行 ngrok -config= ...