J. Bottles

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 ai and bottle volumebi (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 xseconds 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.

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 = 6seconds to do it.

思路：dp

可以先将最小的杯子数求出来。

dp[i][j][k]表示前i个杯子，容量为j时，选了k个的最大剩余水量，然后状态转移看代码；

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<stdlib.h>
  6 #include<queue>
  7 #include<set>
  8 #include<vector>
  9 #include<map>
 10 #include<stack>
 11 #include<deque>
 12 using namespace std;
 13 typedef long long LL;
 14 int ans[1005];
 15 int id[1005];
 16 typedef struct node
 17 {
 18         int x;
 19         int y;
 20         bool operator<(const node&cx)const
 21         {
 22                 if(y == cx.y)return cx.x > x;
 23                 else return y < cx.y;
 24         }
 25 } ss;
 26 priority_queue<ss>que;
 27 bool cmp(node p,node q)
 28 {
 29         return p.y < q.y;
 30 }
 31 ss ak[10005];
 32 int dp[105][10005][105];
 33 int main(void)
 34 {
 35         int n,m;
 36         scanf("%d",&n);
 37         int i,j;
 38         int sum = 0;
 39         int uu ;
 40         for(i = 1; i <= n; i++)
 41         {
 42                 scanf("%d",&ak[i].x);
 43                 sum+=ak[i].x;
 44         }
 45         uu = sum;
 46         int vv = 0;
 47         for(i = 1; i <= n; i++)
 48         {
 49                 scanf("%d",&ak[i].y);
 50                 vv+=ak[i].y;
 51         }
 52         sort(ak+1,ak+n+1,cmp);
 53         int s;
 54         for(i = 0; i < 105; i++)
 55         {
 56                 for(j = 0; j <= 10000; j++)
 57                 {
 58                         for(s = 0; s < 105; s++)
 59                         {
 60                                 dp[i][j][s] = -1e9;
 61                         }
 62                 }
 63         }
 64         for(i = 0; i < 105; i++)
 65         {
 66
 67                         dp[i][0][0]=0;
 68         }
 69         int cn = 0;
 70         int nn = n;
 71         while(sum  > 0)
 72         {
 73                 sum -= ak[nn].y;
 74                 cn++;
 75                 nn--;
 76         }
 77         int maxx = 0;
 78         for(i = 1; i <= n; i++)
 79         {
 80                 for(s = vv; s >= ak[i].y; s--)
 81                 {
 82                         for(j = cn; j >= 1; j--)
 83                         {
 84                                 dp[i][s][j] = max(dp[i][s][j],dp[i-1][s][j]);
 85                                 dp[i][s][j] = max(dp[i][s][j],dp[i-1][s-ak[i].y][j-1]+ak[i].x);
 86                                 if(j == cn&&s >= uu)
 87                                         maxx = max(maxx,dp[i][s][j]);
 88                         }
 89                 }
 90                 for(s = ak[i].y-1; s >= 0; s--)
 91                 {
 92                         for(j = cn ; j >= 1; j--)
 93                         {
 94                                 dp[i][s][j] =max(dp[i-1][s][j],dp[i][s][j]);
 95                                 if(j == cn&&s >= uu)
 96                                         maxx = max(maxx,dp[i][s][j]);
 97                         }
 98                 }
 99         }
100         printf("%d %d\n",cn,uu-maxx);
101         return 0;
102 }

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