csuoj 1391: Boiling Vegetables

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1391

1391: Boiling Vegetables

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 27  Solved: 14
[Submit][Status][Web Board]

Description

Input

Output

Sample Input

0.99 3
2000 3000 4000

Sample Output

6

HINT

Source

分析：

暴力枚举。

AC代码；

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>

using namespace std;

double vagetable[1100],limit;
int n;

int main()
{
    cin>>limit>>n;
    for(int i=0;i<n;i++) cin>>vagetable[i];

    if(n==1)
    {
        cout<<0<<endl;
        return 0;
    }

    sort(vagetable,vagetable+n,less<double>());

    int ans=501;

    for(int i=0;i<n;i++)
    {
        for(int c=0;c<ans;c++)
        {
            double maxsize=vagetable[i]/(c+1);
            double minsize=maxsize*limit;

            int temp=c;
            bool flag=true;

            for(int j=0;j<n;j++)
            {
                if(i==j) continue;

                int c1=(int)ceil(vagetable[j]/maxsize)-1;

                if(vagetable[j]/(c1+1) < minsize)
                {
                    flag=false; break;
                }
                temp+=c1;
                if(temp>=ans)
                {
                    flag=false; break;
                }
            }
            if(flag)
            {
                ans=min(ans,temp);
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
double a[1009];
int main()
{
    double T,maxx,minn;
    int n,i,j,k,PP,sum,flag;
    while(~scanf("%lf%d",&T,&n))
    {
        PP=600;
        for(i=0;i<n;i++)
            scanf("%lf",&a[i]);
        for(i=0;i<n;i++)
            for(j=0;j<=500;j++)
            {
                sum=j;flag=0;
                maxx=a[i]*1.0/(j+1);
                minn=maxx*T;
                for(k=0;k<n && flag==0;k++)
                {
                    if(k==i)
                        continue;
                    int t1=(int)((a[k]-1)/maxx);
                    double t2=a[k]*1.0/(t1+1);
                    if(t2<minn)
                        {flag=1;break;}
                    sum+=t1;
                    if(sum>PP)
                        {flag=1;break;}

                }
                if(flag==0)
                    PP=min(PP,sum);
            }
        printf("%d
",PP);
    }
    return 0;
}