A. Mafia

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/348/problem/A

Description

One day n friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other n - 1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the i-th person wants to play ai rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want?

Input

The first line contains integer n (3 ≤ n ≤ 105). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the i-th number in the list is the number of rounds the i-th person wants to play.

Output

In a single line print a single integer — the minimum number of game rounds the friends need to let the i-th person play at least ai rounds.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Sample Input

3
3 2 2

Sample Output

4

HINT

题意

有n个人,在玩一个游戏,游戏表示每局都必须有个管理员参加

告诉你,每个人想当多少局玩家,然后让你求,最少多少局游戏,才能满足题意!

题解:

贪心就好了,类似厨师煮饼那道题一样

max(max_num,sum/(n-1));

代码:

//qscqesze
#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 200001
#define mod 1000000007
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
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;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
}
//************************************************************************************** ll a[maxn];
ll sum=;
ll mx=;
int main()
{
int n=read();
for(int i=;i<=n;i++)
a[i]=read(),sum+=a[i],mx=max(a[i],mx);
n=n-;
ll ans;
ans=sum/n;
if(sum%n!=)
ans++;
cout<<max(mx,ans)<<endl;
}

Codeforces Round #202 (Div. 1) A. Mafia 贪心的更多相关文章

  1. Codeforces Round #202 (Div. 1) A. Mafia 推公式 + 二分答案

    http://codeforces.com/problemset/problem/348/A A. Mafia time limit per test 2 seconds memory limit p ...

  2. Codeforces Round #202 (Div. 2)

    第一题水题但是wa了一发,排队记录下收到的25,50,100,看能不能找零,要注意100可以找25*3 复杂度O(n) 第二题贪心,先找出最小的花费,然后就能得出最长的位数,然后循环对每个位上的数看能 ...

  3. Codeforces Round #202 (Div. 2) B,C,D,E

    贪心 B. Color the Fence time limit per test 2 seconds memory limit per test 256 megabytes input standa ...

  4. Codeforces Round #382 (Div. 2)B. Urbanization 贪心

    B. Urbanization 题目链接 http://codeforces.com/contest/735/problem/B 题面 Local authorities have heard a l ...

  5. Codeforces Round #164 (Div. 2) E. Playlist 贪心+概率dp

    题目链接: http://codeforces.com/problemset/problem/268/E E. Playlist time limit per test 1 secondmemory ...

  6. Codeforces Round #180 (Div. 2) B. Sail 贪心

    B. Sail 题目连接: http://www.codeforces.com/contest/298/problem/B Description The polar bears are going ...

  7. Codeforces Round #192 (Div. 1) A. Purification 贪心

    A. Purification Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/329/probl ...

  8. Codeforces Round #274 (Div. 1) A. Exams 贪心

    A. Exams Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/480/problem/A Des ...

  9. Codeforces Round #374 (Div. 2) B. Passwords 贪心

    B. Passwords 题目连接: http://codeforces.com/contest/721/problem/B Description Vanya is managed to enter ...

随机推荐

  1. 4-Python数据类型之元组-字符串

    目录 1 元组概念 1.1 元祖的特点 1.2 元组的定义 1.3 元组的访问 1.4 元组的查询 2 命名元组 3 字符串 3.1 字符串的基本操作 3.1.1 字符串的访问 3.1.2 字符串的拼 ...

  2. 36 - 网络编程-TCP编程

    目录 1 概述 2 TCP/IP协议基础 3 TCP编程 3.1 通信流程 3.2 构建服务端 3.3 构建客户端 3.4 常用方法 3.4.1 makefile方法 3.5 socket交互 3.4 ...

  3. nginx之日志设置详解

    nginx的日志设置 access_log access_log是服务器记录了哪些用户,哪些页面以及用户浏览器.ip和其他的访问信息:是一种非常详细的记录信息:如果我们不关心谁访问了我们,可以关闭: ...

  4. nginx 服务器篇

    Nginx 服务器类型 1. Web服务器 Web服务器用于提供HTTP(包括HTTPS)的访问,例如Nginx.Apache.IIS等. 2. 应用程序服务器 应用程序服务器能够用于应用程序的运行, ...

  5. 转:google测试分享-SET和TE

    原文:  http://blog.sina.com.cn/s/blog_6cf812be0102vbnb.html 前端时间看了google测试之道,收获了一些,在此总结下并打算写一个系列blog,顺 ...

  6. C++面试总结

    1.多态       C++多态分两种--静态和动态,其中静态联编支持的多态称为编译时多态,包括重载和模板:动态联编支持的多态称为运行时多态,包括 继承和虚函数实现. 多态主要是由虚函数实现的,虚函数 ...

  7. admin组件详解

    admin组件详解 先根据admin组件启动流程复习下django项目启动至请求过来发生的事 1将admin组件注册进app 2django项目启动 3在运行到定制的admin时执行其下面的apps文 ...

  8. Codeforces 822C Hacker, pack your bags!(思维)

    题目大意:给你n个旅券,上面有开始时间l,结束时间r,和花费cost,要求选择两张时间不相交的旅券时间长度相加为x,且要求花费最少. 解题思路:看了大佬的才会写!其实和之前Codeforces 776 ...

  9. Web开发入门知识小总结

    原来是写给 http://www.zhihu.com/question/22689579 的答案,也算是学了一学期web课程后的一点小总结,搬运到博客里存一下吧~ ================== ...

  10. copy深浅拷贝

    我们在很多方法里都看到copy()方法,这是对变量的复制,赋值,下面来看一下实例: 复制方法调用的是copy模块中的方法: import copy copy.copy()         #前拷贝 c ...