Bottles(选择k个物品的01背包)

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.

做法：dp[k][j]表示的是选择k个瓶子，瓶子里本来有j的苏打水，最大可以达到这么多瓶子的总容积

#include 
#include 
#include 
#include 
#include 
#define maxn 106
using namespace std;
struct Soda
{
    int res;
    int v;
}soda[maxn];
bool cmp(const Soda & a,const Soda & b)
{
    return a.v>b.v;
}
int n,m;
int sodaamount;
int dp[105][10010];
int main()
{
    while(scanf("%d",&n)==1){
    sodaamount=0;
    int a;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&soda[i].res);
        sodaamount+=soda[i].res;
    }
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&soda[i].v);
    }
    sort(soda+1,soda+n+1,cmp);
    int cnt=0;
    int  t=0;
    for(cnt=1;cnt<=n;cnt++)
    {
        t+=soda[cnt].v;
        if(t>=sodaamount)break;
    }
    memset(dp,-1,sizeof(dp));
    int ans=0;
    dp[0][0]=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=sodaamount;j>=soda[i].res;j--)
        {
            for(int k=i;k>=1;k--)
            {
                if(dp[k-1][j-soda[i].res]!=-1)
                dp[k][j]=max(dp[k][j],dp[k-1][j-soda[i].res]+soda[i].v);
            }
        }
    }

    for(int i=sodaamount;i>=0;i--)
    {
        if(dp[cnt][i]>=sodaamount)
        {
            ans=sodaamount-i;
            break;
        }
    }
    printf("%d %d\n",cnt,ans);
    }
    return 0;
}