2010-2011 ACM-ICPC, NEERC, Moscow Subregional Contest Problem C. Contest 水题

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;
}