2016-2017 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred) J dp 背包

J. Bottles

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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

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, ..., a_n (1 ≤ a_i ≤ 100), where a_i is the amount of soda remaining in the i-th bottle.

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

It is guaranteed that a_i ≤ b_i 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.

Examples

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.

题意：给你n个杯子 给出n个杯子的初始水的体积以及杯子的体积 现在倒水 使得用最少个数的杯子装所有的水 并且在倒水的过程中 使得倒出的水尽量的少

题解：爆搜TLE  背包处理 orzzz 太菜了

dp[i][j]表示用i个杯子容量为j最多能装多少水 注意这个j的范围为杯子的体积和 （可能所选的i个杯子 就算装有全部的水 也不会装满）

 /******************************
 code by drizzle
 blog: www.cnblogs.com/hsd-/
 ^ ^    ^ ^
  O      O
 ******************************/
 #include<bits/stdc++.h>
 #include<map>
 #include<set>
 #include<cmath>
 #include<queue>
 #include<bitset>
 #include<math.h>
 #include<vector>
 #include<string>
 #include<stdio.h>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #pragma comment(linker, "/STACK:102400000,102400000")
 using namespace std;
 #define  A first
 #define B second
 const int mod=;
 const int MOD1=;
 const int MOD2=;
 const double EPS=0.00000001;
 typedef __int64 ll;
 const ll MOD=;
 const int INF=;
 const ll MAX=1ll<<;
 const double eps=1e-;
 const double inf=~0u>>;
 const double pi=acos(-1.0);
 typedef double db;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 int n;
 struct node
 {
     int a,b;
 }N[];
 int dp[][];
 bool cmp(struct node aa,struct node bb)
 {
     if(aa.b>bb.b)
         return true;
     if(aa.b==bb.b)
     {
         if(aa.a>bb.a)
             return true;
     }
     return false;
 }
 int main()
 {
     scanf("%d",&n);
     int m=;
     int mm;
     int mm2=;
     for(int i=;i<=n;i++)
     {
         scanf("%d",&N[i].a);
         m+=N[i].a;
     }
     mm=m;
     for(int i=;i<=n;i++)
     {
         scanf("%d",&N[i].b);
         mm2+=N[i].b;
     }
     sort(N+,N++n,cmp);
     int num=;
     while()
     {
         m-=N[++num].b;
         if(m<=)
             break;
     }
     memset(dp,,sizeof(dp));
     for(int i=;i<=num;i++)
      for(int j=;j<=mm2;j++)
         dp[i][j]=-INF;
     dp[][]=;
     for(int j=;j<=n;j++)
     {
         for(int k=mm2;k>=N[j].b;k--)
         {
             for(int l=;l<=num;l++){
                 dp[l][k]=max(dp[l][k],dp[l-][k-N[j].b]+N[j].a);
                 }
         }
     }
     int ans=;
     for(int i=mm2;i>=mm;i--)
         if(dp[num][i])
         ans=max(ans,dp[num][i]);
     printf("%d %d\n",num,mm-ans);
     return ;
 }
 /*
 10
 18 42 5 1 26 8 40 34 8 29
 18 71 21 67 38 13 99 37 47 76
 3 100
 */