Bottles CodeForces - 730J

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Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai?≤?bi).

Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.

Nick asks you to help him to determine k — the minimal number of bottles to store all remaining soda and t — the minimal time to pour soda into k bottles. A bottle can‘t store more soda than its volume. All remaining soda should be saved.

Input

The first line contains positive integer n (1?≤?n?≤?100) — the number of bottles.

The second line contains n positive integers a1,?a2,?...,?an (1?≤?ai?≤?100), where ai is the amount of soda remaining in the i-th bottle.

The third line contains n positive integers b1,?b2,?...,?bn (1?≤?bi?≤?100), where bi is the volume of the i-th bottle.

It is guaranteed that ai?≤?bi for any i.

Output

The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.

Example

Input
4
3 3 4 3
4 7 6 5
Output
2 6
Input
2
1 1
100 100
Output
1 1
Input
5
10 30 5 6 24
10 41 7 8 24
Output
3 11

Note

In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3?+?3?=?6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1?+?2?=?3 seconds. So, all the soda will be in two bottles and he will spend 3?+?3?=?6 seconds to do it.

题解:dp[ i ][ j ][ k ]表示1~i个瓶子中已选j个瓶子得到的最大实际装水量。那么dp[ i ][ j ][ k ]=max(dp[ i ] [ j ] [ k ],dp[ i - 1 ][ j - 1][ k - b[ i ] ] + a[ i ] )。

首先,题目要求最少的瓶子最短的时间,而最少的瓶子满足其总容量大于->实际水的总量<-(sumN)。所需要的时间,则是水的总量(sumN)减去这k个瓶子装的水的总量。所以time=sum1-maxSum(K)。

dp好难,初始化都不好掌握 ,><

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int maxn=105;
 5 
 6 int n,sum1,sum2;
 7 int dp[maxn][maxn*maxn];
 8 
 9 struct node{
10     int a,b;
11     bool operator<(const node& i)const{
12         return i.b<b;
13     }
14 }p[maxn];
15 
16 void Inite(){
17     memset(dp,-1,sizeof(dp));
18     dp[0][0]=0;
19 }
20 
21 void Solve(){
22     int tem=0,q=1;
23     for(int i=1;i<=n;i++){
24         tem+=p[i].b;
25         q=i;
26         if(tem>=sum1) break;
27     }
28     for(int i=1;i<=n;i++)
29         for(int j=q;j>=1;j--)
30             for(int k=sum2;k>=p[i].b;k--) 
31                 if(~dp[j-1][k-p[i].b]) dp[j][k]=max(dp[j][k],dp[j-1][k-p[i].b]+p[i].a);
32     int ans=0;
33     for(int i=sum1;i<=sum2;i++) ans=max(ans,dp[q][i]);
34     cout<<q<<" "<<sum1-ans<<endl;
35 }
36 
37 int main()
38 {    
39     cin>>n;
40     Inite();
41     sum1=sum2=0;
42     for(int i=1;i<=n;i++){ cin>>p[i].a; sum1+=p[i].a; }
43     for(int i=1;i<=n;i++){ cin>>p[i].b; sum2+=p[i].b; }
44     sort(p+1,p+n+1);
45     Solve();
46 }

 

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