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；

 

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <cmath>
 6 
 7 using namespace std;
 8 
 9 double vagetable[1100],limit;
10 int n;
11 
12 int main()
13 {
14     cin>>limit>>n;
15     for(int i=0;i<n;i++) cin>>vagetable[i];
16 
17     if(n==1)
18     {
19         cout<<0<<endl;
20         return 0;
21     }
22 
23     sort(vagetable,vagetable+n,less<double>());
24 
25     int ans=501;
26 
27     for(int i=0;i<n;i++)
28     {
29         for(int c=0;c<ans;c++)
30         {
31             double maxsize=vagetable[i]/(c+1);
32             double minsize=maxsize*limit;
33 
34             int temp=c;
35             bool flag=true;
36 
37             for(int j=0;j<n;j++)
38             {
39                 if(i==j) continue;
40 
41                 int c1=(int)ceil(vagetable[j]/maxsize)-1;
42 
43                 if(vagetable[j]/(c1+1) < minsize)
44                 {
45                     flag=false; break;
46                 }
47                 temp+=c1;
48                 if(temp>=ans)
49                 {
50                     flag=false; break;
51                 }
52             }
53             if(flag)
54             {
55                 ans=min(ans,temp);
56             }
57         }
58     }
59     cout<<ans<<endl;
60     return 0;
61 }

View Code

 

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<math.h>
 4 #include<iostream>
 5 using namespace std;
 6 double a[1009];
 7 int main()
 8 {
 9     double T,maxx,minn;
10     int n,i,j,k,PP,sum,flag;
11     while(~scanf("%lf%d",&T,&n))
12     {
13         PP=600;
14         for(i=0;i<n;i++)
15             scanf("%lf",&a[i]);
16         for(i=0;i<n;i++)
17             for(j=0;j<=500;j++)
18             {
19                 sum=j;flag=0;
20                 maxx=a[i]*1.0/(j+1);
21                 minn=maxx*T;
22                 for(k=0;k<n && flag==0;k++)
23                 {
24                     if(k==i)
25                         continue;
26                     int t1=(int)((a[k]-1)/maxx);
27                     double t2=a[k]*1.0/(t1+1);
28                     if(t2<minn)
29                         {flag=1;break;}
30                     sum+=t1;
31                     if(sum>PP)
32                         {flag=1;break;}
33 
34                 }
35                 if(flag==0)
36                     PP=min(PP,sum);
37             }
38         printf("%d\n",PP);
39     }
40     return 0;
41 }

View Code