Codeforces gym 100685 C. Cinderella 水题
C. Cinderella
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100685/problem/C
Description
Cinderella is given a task by her Stepmother before she is allowed to go to the Ball. There are N (1 ≤ N ≤ 1000) bottles with water in the kitchen. Each bottle contains Li (0 ≤ Li ≤ 106) ounces of water and the maximum capacity of each is 109 ounces. To complete the task Cinderella has to pour the water between the bottles to fill them at equal measure.
Cinderella asks Fairy godmother to help her. At each turn Cinderella points out one of the bottles. This is the source bottle. Then she selects any number of other bottles and for each bottle specifies the amount of water to be poured from the source bottle to it. Then Fairy godmother performs the transfusion instantly.
Please calculate how many turns Cinderella needs to complete the Stepmother's task.
Input
The first line of input contains an integer number N (1 ≤ N ≤ 1000) — the total number of bottles.
On the next line integer numbers Li are contained (0 ≤ Li ≤ 106) — the initial amount of water contained in ith bottle.
Output
Output a single line with an integer S — the minimal number of turns Cinderella needs to complete her task.
Sample Input
3
5 7 7
Sample Output
2
HINT
题意
每一个回合可以选择一瓶水然后给其他杯子倒水,然后问你最少几个回合可以使得所有的杯子里面的水都一样
题解:
求一个平均数,然后大于平均数的就直接倒水就好了
代码
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 2000001
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************** ll sum=;
ll a[maxn];
int main()
{
int n=read();
for(int i=;i<n;i++)
a[i]=read(),sum+=a[i];
ll kiss=sum/n;
int ans=;
for(int i=;i<n;i++)
if(a[i]>kiss)
ans++;
cout<<ans<<endl;
}
Codeforces gym 100685 C. Cinderella 水题的更多相关文章
- Codeforces Gym 100286G Giant Screen 水题
Problem G.Giant ScreenTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/con ...
- Codeforces Gym 100431D Bubble Sort 水题乱搞
原题链接:http://codeforces.com/gym/100431/attachments/download/2421/20092010-winter-petrozavodsk-camp-an ...
- Codeforces GYM 100114 B. Island 水题
B. Island Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Description O ...
- Codeforces gym 100685 A. Ariel 暴力
A. ArielTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100685/problem/A Desc ...
- Codeforces gym 100685 F. Flood bfs
F. FloodTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100685/problem/F Desc ...
- Codeforces gym 100685 E. Epic Fail of a Genie 贪心
E. Epic Fail of a GenieTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100685 ...
- codeforces 577B B. Modulo Sum(水题)
题目链接: B. Modulo Sum time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #367 (Div. 2)---水题 | dp | 01字典树
A.Beru-taxi 水题:有一个人站在(sx,sy)的位置,有n辆出租车,正向这个人匀速赶来,每个出租车的位置是(xi, yi) 速度是 Vi;求人最少需要等的时间: 单间循环即可: #inclu ...
- codeforces 696A Lorenzo Von Matterhorn 水题
这题一眼看就是水题,map随便计 然后我之所以发这个题解,是因为我用了log2()这个函数判断在哪一层 我只能说我真是太傻逼了,这个函数以前听人说有精度问题,还慢,为了图快用的,没想到被坑惨了,以后尽 ...
随机推荐
- TCP/IP详解学习笔记(8)-DNS域名系统
前面已经提到了访问一台机器要靠IP地址和MAC地址,其中,MAC地址可以通过ARP协议得到,所以这对用户是透明的,但是IP地址就不行,无论如何用户都需要用一个指定的IP来访问一台计算机,而IP地址又非 ...
- information_schema中的三个关于锁的表
在5.5中,information_schema 库中增加了三个关于锁的表(MEMORY引擎):innodb_trx ## 当前运行的所有事务innodb_locks ## ...
- [Papers]NSE, $\p_3u$, Lebesgue space [Kukavica-Ziane, JMP, 2007]
$$\bex \p_3\bbu\in L^p(0,T;L^q(\bbR^3)),\quad \frac{2}{p}+\frac{3}{q}=2,\quad \frac{9}{4}\leq q\leq ...
- Matlab编程实例(2) 同期平均
%多点同期平均 close all; clear all; pi = 3.14159; Samp2=input('您需要几组信号做同期平均?') Samp1=1000 %设置采样精度 t = lins ...
- HDU5779 Tower Defence (BestCoder Round #85 D) 计数dp
分析(官方题解): 一点感想:(这个题是看题解并不是特别会转移,当然写完之后看起来题解说得很清晰,主要是人太弱 这个题是参考faebdc神的代码写的,说句题外话,很荣幸高中和faebdc巨一个省,虽然 ...
- Rust 中的继承与代码复用
在学习Rust过程中突然想到怎么实现继承,特别是用于代码复用的继承,于是在网上查了查,发现不是那么简单的. C++的继承 首先看看c++中是如何做的. 例如要做一个场景结点的Node类和一个Sprit ...
- ORA-12162: TNS:net service name is incorrectly specified
今天在进行修改oracle_sid环境变量的时候,将相关的环境变量值去掉,从而不能进入sqlplus,报错如下: [oracle@kel ~]$ sqlplus / as sysdba SQL*Plu ...
- python中List的sort方法的用法
python列表排序 简单记一下python中List的sort方法(或者sorted内建函数)的用法. 关键字: python列表排序 python字典排序 sorted List的元素可以是各种东 ...
- 连分数(分数类模板) uva6875
//连分数(分数类模板) uva6875 // 题意:告诉你连分数的定义.求连分数,并逆向表示出来 // 思路:直接上分数类模板.要注意ai可以小于0 #include <iostream> ...
- [算法] 插入排序 Insertion Sort
插入排序(Insertion Sort)是一种简单直观的排序算法.它的工作原理是通过构建有序序列,对于未排序数据,在已排序序列中从后向前扫描,找到相应位置并插入.插入排序在实现上,通常采用in-pla ...