Codeforces Round #202 (Div. 1) A. Mafia 贪心
A. Mafia
Time Limit: 20 Sec Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/348/problem/A
Description
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
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- 深入理解C指针----学习笔记
深入理解C指针 第1章 认识指针 理解指针的关键在于理解C程序如何管理内存,指针包含的就是内存地址. 1.1 指针和内存 C程序在编译后,以三种方式使用内存: 1. 静态. ...
- pythonif语句和循环语句
1.if语句用法 # if语句用法(缩进相同的成为一个代码块) score=90 if score>=60: print("合格") print("OK" ...
- udhcpc命令
要使用网络通讯,所以不可避免的要用到dhcp.理想的网络通讯方式是下面3种都要支持: 1,接入已有网络.这便要求可以作为dhcp客户端. 2,作为DHCP服务器,动态分配IP. 简单说下前2种情况. ...
- linux device tree源代码解析--转
//Based on Linux v3.14 source code Linux设备树机制(Device Tree) 一.描述 ARM Device Tree起源于OpenFirmware (OF), ...
- oracle命令生成AWR报告
--命令生成AWR报告oracle@linux:~> sqlplus / as sysdba SQL*Plus: Release 11.1.0.7.0 - Production on Fri A ...
- 20165301 预备作业三:Linux安装及命令入门
预备作业三:Linux安装及命令入门 VirtualBox虚拟机的安装 在进行安装之前,原本以为有了娄老师的安装教程会是一件很容易的事情.万万没想到,在自己实际动手操作中,还是遇到了许多困难.通过与同 ...
- C语言俄罗斯方块
#include <windows.h> #include <stdio.h> #include <time.h> #include <conio.h> ...
- LeetCode765. Couples Holding Hands
N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum numb ...
- Java学习笔记()ArrayList
1.什么是ArrayList ArrayList就是传说中的动态数组,用MSDN中的说法,就是Array的复杂版本,它提供了如下一些好处: 动态的增加和减少元素 实现了ICollection和ILis ...
- MySQL的表管理
首先,先选择数据库(极其特别重要,如果不选择,将默认为第一个数据库) mysql > use db_name; 查看所有表 mysql > show tables; 1.创建表 creat ...