Problem C. Contest

题目连接:

http://codeforces.com/gym/100714

Description

The second round of the annual student collegiate programming contest is being held in city N. To

be prepared for the inrush of participants, the jury needs to know the number of them attending the

previous, first round.

Unfortunately, the statistics regarding that first round (including the final standings) was lost during a

recent disk failure and no backup was made.

The only hope is a short statistical summary that was found written on a tiny piece of paper by the oldest

jury member. The percentage of teams which have solved the problem is provided for each problem of

the the first round. Each percentage is an integer rounded using the usual mathematical rules (numbers

with a fractional part strictly less than .5 are rounded down, the others are rounded up).

This is the only information the jury has at hand. Also, that oldest jury member clearly remembers that

a prize was awarded to some team during the first round, probably for winning it. Hence, at least one

team had participated in the first round.

Input

The first line of input contains an integer N (3 ≤ N ≤ 12) — the total number of problems in the

contest. The second line of input contains N integers P1, . . . , PN . Each number Pi (0 ≤ Pi ≤ 100)

denotes a percentage of the teams solved the i

th problem.

Output

Print out the minimum possible number of teams that could have participated in the first round.

Sample Input

3

33 67 100

Sample Output

3

Hint

题意

告诉你每道题的过题率,但是都是四舍五入了的。

请你输出最少有多少个队伍参加,才能满足这个过题率。

至少为1个队

题解:

数据范围很小嘛,就暴力枚举人数就好了。

代码

#include <bits/stdc++.h>

using namespace std;

int v[110][110],a[110],n;

int main()
{
//freopen("out.txt","w",stdout);
for(int i=1;i<=100;i++)
{
v[i][0]=1;
for(int j=1;j<=i;j++)
{
double x=j/((double)1.0*i)*(double)100.0;
int y=(int)(x+(double)0.5000000000001);
v[i][y]=1;
}
}
/* for(int i = 1 ; i <= 100 ; ++ i){
for(int j = 1 ; j <= i ; ++ j) printf("%d,%d,%d\n" , i , j , v[i][j]);
}*/
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
for(int i=1;i<=100;i++)
{
int flag=1;
for(int j=1;j<=n;j++)
if(!v[i][a[j]])
{
flag=0;
break;
}
if(flag)
{
printf("%d\n",i);
return 0;
}
}
return 0;
}

2010-2011 ACM-ICPC, NEERC, Moscow Subregional Contest Problem C. Contest 水题的更多相关文章

  1. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 G. Garden Gathering

    Problem G. Garden Gathering Input file: standard input Output file: standard output Time limit: 3 se ...

  2. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 D. Delay Time

    Problem D. Delay Time Input file: standard input Output file: standard output Time limit: 1 second M ...

  3. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 I. Illegal or Not?

    I. Illegal or Not? time limit per test 1 second memory limit per test 512 megabytes input standard i ...

  4. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 K. King’s Rout

    K. King's Rout time limit per test 4 seconds memory limit per test 512 megabytes input standard inpu ...

  5. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 H. Hashing

    H. Hashing time limit per test 1 second memory limit per test 512 megabytes input standard input out ...

  6. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 C. Colder-Hotter

    C. Colder-Hotter time limit per test 1 second memory limit per test 512 megabytes input standard inp ...

  7. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 A. Anagrams

    A. Anagrams time limit per test 1 second memory limit per test 512 megabytes input standard input ou ...

  8. 2010-2011 ACM-ICPC, NEERC, Moscow Subregional Contest Problem J. Joke 水题

    Problem J. Joke 题目连接: http://codeforces.com/gym/100714 Description The problem is to cut the largest ...

  9. 2010-2011 ACM-ICPC, NEERC, Moscow Subregional Contest Problem H. Hometask 水题

    Problem H. Hometask 题目连接: http://codeforces.com/gym/100714 Description Kolya is still trying to pass ...

  10. 2018-2019 ICPC, NEERC, Southern Subregional Contest

    目录 2018-2019 ICPC, NEERC, Southern Subregional Contest (Codeforces 1070) A.Find a Number(BFS) C.Clou ...

随机推荐

  1. Codeforces 666 B. World Tour

    http://codeforces.com/problemset/problem/666/B 题意: 给定一张边权均为1的有向图,求四个不同的点A,B,C,D,使得dis[A][B]+dis[B][C ...

  2. Does Deep Learning Come from the Devil?

    Does Deep Learning Come from the Devil? Deep learning has revolutionized computer vision and natural ...

  3. 【原创】backbone1.1.0源码解析之View

    作为MVC框架,M(odel)  V(iew)  C(ontroler)之间的联系是必不可少的,今天要说的就是View(视图) 通常我们在写逻辑代码也好或者是在ui组件也好,都需要跟dom打交道,我们 ...

  4. Javascript加速运动与减速运动

    加速运动,即一个物体运动时速度越来越快:减速运动,即一个物体运动时速度越来越慢.现在用Javascript来模拟这两个效果,原理就是用setInterval或setTimeout动态改变一个元素与另外 ...

  5. SQL SERVER C#数据库操作类(连接、执行SQL)

    using System; using System.Collections; using System.Collections.Specialized; using System.Data; usi ...

  6. FPGA学习笔记. DDS

    DDS原理 直接数字式频率合成器(Direct Digital Synthesizer) 频率计算公式 Fout = FW * Fclk / 2^N Fout 输出频率, Fw 频率控制字, N 位数 ...

  7. F - A计划

    题目链接: https://cn.vjudge.net/contest/254150#problem/F wa代码: #include<iostream> #include<stri ...

  8. Linux下可以使用ps命令来查看Oracle相关的进程

    Linux下可以使用ps命令来查看Oracle相关的进程 Oracle Listener 这个命令会列出Oracle Net Listener的进程 [oracle@ www.linuxidc.com ...

  9. Visual Studio 2012“完美的拥抱”Visual Studio Online

    看了Visual Studio 2012完美的拥抱GitHub 写的不错,不过,配置起来太麻烦.既然是使用VS编码,微软的东西嘛,当然还有更简单的,那就是Visual Studio Online.不用 ...

  10. 设置adb server的端口号

    在操作系统的系统环境里面,加一个环境变量: ANDROID_ADB_SERVER_PORT,值为9999,看自己喜欢.