Candy Sharing Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4425    Accepted Submission(s): 2698

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

题解:深搜一下。。。关键要开两个数组;因为自己给下一个一半的时候自己变了,所以再开个数组记录上一状态;

代码:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define P_ printf(" ")
#define T_T while(T--)
typedef long long LL;
const int INF=0x3f3f3f3f;
int N;
int m[110],n[110];
int ans;
void dfs(int t){
if(ans)return;
if(*max_element(m,m+N)==*min_element(m,m+N)){
ans=1;
printf("%d %d\n",t,m[0]);
return ;
}
for(int i=1;i<N;i++){
m[i]=(n[i-1]+n[i])/2;
if(m[i]&1)m[i]++;
// if(m[i]&1)m[i]++;
}
m[0]=(n[0]+n[N-1])/2;
if(m[0]&1)m[0]++;
for(int i=0;i<N;i++)n[i]=m[i];
// if(m[0]&1)m[0]++;
//for(int i=0;i<N;i++)printf("%d ",m[i]);puts("");
//getchar();
dfs(t+1);
}
int main(){
while(SI(N),N!=0){
for(int i=0;i<N;i++)scanf("%d",&n[i]),m[i]=n[i];
ans=0;
dfs(0);
}
return 0;
}

  

Candy Sharing Game(模拟搜索)的更多相关文章

  1. 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 ...

  2. 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> ...

  3. 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 ...

  4. hdu 1034 Candy Sharing Game

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

  5. 【LOJ6254】最优卡组 堆(模拟搜索)

    [LOJ6254]最优卡组 题面 题解:常用的用堆模拟搜索套路(当然也可以二分).先将每个卡包里的卡从大到小排序,然后将所有卡包按(最大值-次大值)从小到大排序,并提前处理掉只有一张卡的卡包. 我们将 ...

  6. 【BZOJ4524】[Cqoi2016]伪光滑数 堆(模拟搜索)

    [BZOJ4524][Cqoi2016]伪光滑数 Description 若一个大于1的整数M的质因数分解有k项,其最大的质因子为Ak,并且满足Ak^K<=N,Ak<128,我们就称整数M ...

  7. 【BZOJ4345】[POI2016]Korale 堆(模拟搜索)

    [BZOJ4345][POI2016]Korale Description 有n个带标号的珠子,第i个珠子的价值为a[i].现在你可以选择若干个珠子组成项链(也可以一个都不选),项链的价值为所有珠子的 ...

  8. JavaScript在表格中模拟搜索多关键词搜索和筛选

    模拟搜索需要实现以下功能: 1.用户的模糊搜索不区分大小写,需要小写字母匹配同样可以匹配到该字母的大写单词. 2.多关键词模糊搜索,假设用户关键词以空格分隔,在关键词不完整的情况下仍然可以匹配到包含该 ...

  9. 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 ...

随机推荐

  1. Android应用如何开机自启动、自启动失败原因

    本文主要介绍Android应用如何开机自启动.自启动失败的原因.adb命令发送BOOT_COMPLETED.问题:应用程序是否可以在安装后自启动,没有ui的纯service应用如何启动?答案马上揭晓^ ...

  2. BZOJ 4000: [TJOI2015]棋盘( 状压dp + 矩阵快速幂 )

    状压dp, 然后转移都是一样的, 矩阵乘法+快速幂就行啦. O(logN*2^(3m)) ------------------------------------------------------- ...

  3. BZOJ 3439: Kpm的MC密码( trie + DFS序 + 主席树 )

    把串倒过来插进trie上, 那么一个串的kpm串就是在以这个串最后一个为根的子树, 子树k大值的经典问题用dfs序+可持久化线段树就可以O(NlogN)解决 --------------------- ...

  4. 一周学会Mootools 1.4中文教程:(3)事件

    今天我們講解一下mt的事件部分,对于事件的讲解主要包含三部分,分别是:绑定,移除,和触发,我们首先来看一个例子 //jquery的事件绑定方式$('a').click(function){ alert ...

  5. expect交互式自动化脚本

    一 什么是expect 1 Expect is a tool for automating interactive applications such as telnet, ftp, passwd, ...

  6. python strip()函数介绍

    函数原型 声明:str为字符串,s为要删除的字符序列 str.strip(s)        删除str字符串中开头.结尾处,位于 s删除序列的字符 str.lstrip(s)       删除str ...

  7. 搭建Ubuntu环境中的Error [dpkg 被中断,您必须手工运行 sudo dpkg --configure -a 解决此问题][安装Flashplayer出错 ]

    //解决方法如下: sudo rm /var/cache/apt/archives/lock sudo rm /var/lib/dpkg/lock sudo dpkg -r flashplugin-i ...

  8. BBM(Bad Block Management)坏块管理

    不管WL算法如何高明,在使用中都会碰到一个头痛的问题,那就是坏块,所以一个SSD必须要有坏块管理机制.何谓坏块?一个闪存块里包含有不稳定的地址,不能保证读/写/擦时数据的准确性.            ...

  9. 一个简单的反射连接程序(修改文件时间,以及创建Windows服务)

    program SvrDemo; uses  Windows,  WinSvc,  winsock; const  RegName = 'SvrDemo'; var  szServiceName: p ...

  10. VS2010/MFC常用控件:图片控件Picture Control

    图片控件Picture Control 本节主要讲一种简单实用的控件,图片控件Picture Control.我们可以在界面某个位置放入图片控件,显示图片以美化界面. 图片控件简介 图片控件和前面讲到 ...