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. poj 3252 Round Numbers 数位dp

    题目链接 找一个范围内二进制中0的个数大于等于1的个数的数的数量.基础的数位dp #include<bits/stdc++.h> using namespace std; #define ...

  2. win7中注册tomcat服务

    非安装版tomcat下载后,在bin文件夹会有一个startup.bat文件,运行该文件即可启动tomcat了.不过在服务器配置tomcat的话,就通常需要注册为服务. 在/bin文件下还有tomca ...

  3. XSS CSRF

    XSS CSRF XSS 参考 https://zh.wikipedia.org/wiki/%E8%B7%A8%E7%B6%B2%E7%AB%99%E6%8C%87%E4%BB%A4%E7%A2%BC ...

  4. 解析Tensorflow官方PTB模型的demo

    RNN 模型作为一个可以学习时间序列的模型被认为是深度学习中比较重要的一类模型.在Tensorflow的官方教程中,有两个与之相关的模型被实现出来.第一个模型是围绕着Zaremba的论文Recurre ...

  5. 【Delphi内联汇编学习1】Delphi与汇编

    我一直认为Delphi功能与C++相比毫不逊色,提供了丰富的控件和类.全部API以及嵌入的汇编.最近小弟在把C版的Huffman压缩改用Delphi写时,顺便“研究”了一下Delphi的位操作和嵌入式 ...

  6. iOS搜索指定字符在字符串中的位置

    NSString *tmpStr = @"asd341234aaaaccd"; NSRange range; range = [tmpStr rangeOfString:@&quo ...

  7. java设计模式(二)单例模式 建造者模式

    (三)单例模式 单例模式应该是最常见的设计模式,作用是保证在JVM中,该对象仅仅有一个实例存在. 长处:1.降低某些创建比較频繁的或者比較大型的对象的系统开销. 2.省去了new操作符,减少系统内存使 ...

  8. Android 架构【转】

    import java.util.ArrayList; import java.util.List;   import android.app.Activity; import android.app ...

  9. (译)Node.js的 EventEmitter 教程

    原文标题:Node.js EventEmitter Tutorial 原文链接:http://www.hacksparrow.com/node-js-eventemitter-tutorial.htm ...

  10. android http同步请求

    1.界面 <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" xmlns:too ...