Bottles CodeForces - 730J (limiting backpack)

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.

Meaning of the questions:
choose as little pond, water as large as possible, and volume enough to hold all the water.
Ideas:
a prerequisite pool as little as possible, then it is surely the selection of a few greedy large pool, the number of pools can be directly determined.

Definition of F [i] [k] [j], i represents a pool before the k selected for the tank volume to maximum amount of water j.
According to the backpack transferred just fine.

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 105;
int f[2][105][10005];
int a[maxn],b[maxn],tb[maxn],sum[maxn];

int cmp(int a,int b)
{
    return a > b;
}

int main()
{
    int n;scanf("%d",&n);
    int num = 0;
    for(int i = 1;i <= n;i++)
    {
        scanf("%d",&a[i]);
        num += a[i];
    }
    for(int i = 1;i <= n;i++)
    {
        scanf("%d",&b[i]);
        tb[i] = b[i];
        sum[i] = sum[i - 1] + b[i];
    }
    sort(b + 1,b + 1 + n,cmp);
    int K = 1;
    int tsum = 0;
    for(;K <= n;K++)
    {
        tsum += b[K];
        if(tsum >= num)break;
    }
    
    memset(f,-INF,sizeof(f));
    
    f[0 & 1][0][0] = f[1 & 1][0][0] = 0;
    for(int i = 1;i <= n;i++)
    {
        for(int j = 0;j <= sum[i];j++)
        {
            for(int k = 1;k <= K && k <= i;k++)
            {
                f[i & 1][k][j] = f[(i - 1) & 1][k][j];
                if(j >= tb[i])
                {
                    f[i & 1][k][j] = max(f[i & 1][k][j],f[(i - 1) & 1][k - 1][j - tb[i]] + a[i]);
                }
            }
        }
    }
    
    int ans = -INF;
    for(int i = num;i <= sum[n];i++)
    {
        ans = max(ans,f[n & 1][K][i]);
    }
    printf("%d %d\n",K,num - ans);
    return 0;
}