sicily 1052. Candy Sharing Game
Constraints
Time Limit: 1 secs, Memory Limit: 32 MB
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 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
#include <iostream>
#include <vector> using namespace std; int main(int argc, char const *argv[])
{
int stuNum;
vector<int> stuCandy;
while (cin >> stuNum && stuNum != ) {
stuCandy.resize(stuNum);
for (int i = ; i != stuNum; ++i)
cin >> stuCandy[i];
int roundNum = ;
while (++roundNum) {
int firstCandy = stuCandy[];
int i;
for (i = ; i != stuNum - ; ++i) {
stuCandy[i] -= (stuCandy[i] / );
stuCandy[i] += (stuCandy[i + ] / );
if (stuCandy[i] % != )
stuCandy[i] += ;
}
stuCandy[i] -= (stuCandy[i] / );
stuCandy[i] += (firstCandy / );
if (stuCandy[i] % != )
stuCandy[i] += ; for (i = ; i != stuNum - ; ++i) {
if (stuCandy[i] != stuCandy[i + ])
break;
}
if (i == stuNum - )
break;
}
cout << roundNum << " " << stuCandy[] << endl;
}
return ;
}
sicily 1052. Candy Sharing Game的更多相关文章
- 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 ...
- 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 ...
- 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 ...
- 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> ...
随机推荐
- BZOJ 1407 Savage(拓展欧几里得)
这题的时间复杂度真玄学... O(m*n^2).1e8也能过啊... 首先题目保证m<=1e6. 这启发我们枚举或者二分答案? 但是答案不满足单调性,考虑从小到大枚举m. 对于每一个m,枚举两个 ...
- BZOJ4565 HAOI2016字符合并(区间dp+状压dp)
设f[i][j][k]为将i~j的字符最终合并成k的答案.转移时只考虑最后一个字符是由哪段后缀合成的.如果最后合成为一个字符特殊转移一下. 复杂度看起来是O(n32k),实际常数极小达到O(玄学). ...
- bzoj 1877: [SDOI2009]晨跑 (网络流)
明显拆点费用流: type arr=record toward,next,cap,cost:longint; end; const mm=<<; maxn=; maxm=; var edg ...
- hdu1950 Bridging signals
LIS nlogn的时间复杂度,之前没有写过. 思路是d[i]保存长度为i的单调不下降子序列末尾的最小值. 更新时候,如果a[i]>d[len],(len为目前最长的单调不下降子序列) d[++ ...
- [Leetcode] Sum root to leaf numbers求根到叶节点的数字之和
Given a binary tree containing digits from0-9only, each root-to-leaf path could represent a number. ...
- caffe数据集——LMDB
LMDB介紹 Caffe使用LMDB來存放訓練/測試用的數據集,以及使用網絡提取出的feature(為了方便,以下還是統稱數據集).數據集的結構很簡單,就是大量的矩陣/向量數據平鋪開來.數據之間沒有什 ...
- Coconuts HDU - 5925 二维离散化 自闭了
TanBig, a friend of Mr. Frog, likes eating very much, so he always has dreams about eating. One day, ...
- 转ajax的jsonp的文章
转:http://justcoding.iteye.com/blog/1366102/ Js是不能跨域请求.出于安全考虑,js设计时不可以跨域. 什么是跨域: 1.域名不同时. 2.域名相同,端口不同 ...
- POJ 3421分解质因数
X-factor Chains Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7375 Accepted: 2340 D ...
- matlab求一个矩阵中各元素出现的个数(归一化)
function [m,n] = stamatrix(a) %网上找到的方法,感觉很巧妙 x=a(:); x=sort(x); d=diff([x;max(x)+1]); count = diff(f ...