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[],limit;

 int n;

 int main()

 {

     cin>>limit>>n;

     for(int i=;i<n;i++) cin>>vagetable[i];

     if(n==)

     {

         cout<<<<endl;

         return ;

     }

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

     int ans=;

     for(int i=;i<n;i++)

     {

         for(int c=;c<ans;c++)

         {

             double maxsize=vagetable[i]/(c+);

             double minsize=maxsize*limit;

             int temp=c;

             bool flag=true;

             for(int j=;j<n;j++)

             {

                 if(i==j) continue;

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

                 if(vagetable[j]/(c1+) < minsize)

                 {

                     flag=false; break;

                 }

                 temp+=c1;

                 if(temp>=ans)

                 {

                     flag=false; break;

                 }

             }

             if(flag)

             {

                 ans=min(ans,temp);

             }

         }

     }

     cout<<ans<<endl;

     return ;

 }

 #include<stdio.h>

 #include<string.h>

 #include<math.h>

 #include<iostream>

 using namespace std;

 double a[];

 int main()

 {

     double T,maxx,minn;

     int n,i,j,k,PP,sum,flag;

     while(~scanf("%lf%d",&T,&n))

     {

         PP=;

         for(i=;i<n;i++)

             scanf("%lf",&a[i]);

         for(i=;i<n;i++)

             for(j=;j<=;j++)

             {

                 sum=j;flag=;

                 maxx=a[i]*1.0/(j+);

                 minn=maxx*T;

                 for(k=;k<n && flag==;k++)

                 {

                     if(k==i)

                         continue;

                     int t1=(int)((a[k]-)/maxx);

                     double t2=a[k]*1.0/(t1+);

                     if(t2<minn)

                         {flag=;break;}

                     sum+=t1;

                     if(sum>PP)

                         {flag=;break;}

                 }

                 if(flag==)

                     PP=min(PP,sum);

             }

         printf("%d\n",PP);

     }

     return ;

 }