2016-2017 ACM-ICPC, NEERC, Southern Subregional Contest 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 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.

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.

               

dp[i][j]表示选了前i个瓶子容量为j时能装下的最多水量。

#include <bits/stdc++.h>
using namespace std;
int dp[106][10006]; //选了前i个杯子容量为j时最多能装多少水。 因为要求倒最少的水
struct node
{
int sv,v;
} f[105];
bool cmp(node a,node b)
{
if(a.v!=b.v)
return a.v>b.v;
else
return a.sv>b.sv;
}
int main()
{
int n;
cin>>n;
int sum1,sum2;
sum1=sum2=0;
for(int i=1; i<=n; i++)
{
cin>>f[i].sv;
sum1+=f[i].sv;
}
for(int i=1; i<=n; i++)
{
cin>>f[i].v;
sum2+=f[i].v;
}
int x=sum1;
int num=0;
sort(f+1,f+1+n,cmp);
while(1)
{
x-=f[++num].v;
if(x<=0)
break;
}
for(int i=0; i<=105; i++)
for(int j=0; j<=10005; j++)
dp[i][j]=-999999;
dp[0][0]=0;
for(int i=1; i<=n; i++)
{
for(int j=sum2; j>=f[i].v; j--)
for(int k=1; k<=num; k++)
dp[k][j]=max(dp[k][j],dp[k-1][j-f[i].v]+f[i].sv);
}
int ans=0;
for(int i=sum1; i<=sum2; i++)
{
ans=max(ans,dp[num][i]);
// cout<<dp[num][i]<<endl;
}
printf("%d %d\n",num,sum1-ans);
}