USACO 2013 Nov Gold 3.No Change

Description

Farmer John is at the market to purchase supplies for his farm.  He has in his pocket K coins (1 <= K <= 16), each with value in the range 1..100,000,000.  FJ would like to make a sequence of N purchases (1 <= N <= 100,000), where the ith purchase costs c(i) units of money (1 <= c(i) <= 10,000).  As he makes this sequence of purchases, he can periodically stop and pay, with a single coin, for all the purchases made since his last payment (of course, the single coin he uses must be large enough to pay for all of these).  Unfortunately, the vendors at the market are completely out of change, so whenever FJ uses a coin that is larger
than the amount of money he owes, he sadly receives no changes in return!

Please compute the maximum amount of money FJ can end up with after making his N purchases in sequence.  Output -1 if it is impossible for FJ to make all of his purchases.

Input

* Line 1: Two integers, K and N.
* Lines 2..1+K: Each line contains the amount of money of one of FJ's coins.
* Lines 2+K..1+N+K: These N lines contain the costs of FJ's intended purchases.

Output

* Line 1: The maximum amount of money FJ can end up with, or -1 if FJ cannot complete all of his purchases.

Sample Input

3 6
12
15
10
6
3
3
2
3
7

Sample Output

12


HINT

INPUT DETAILS:

FJ has 3 coins of values 12, 15, and 10.  He must make purchases insequence of value 6, 3, 3, 2, 3, and 7.

OUTPUT DETAILS:

FJ spends his 10-unit coin on the first two purchases, then the 15-unit coin on the remaining purchases.  This leaves him with the 12-unit coin.

题目分析：

f[s]表示用S状态的硬币能买到的最多货物

#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
int n,m;
long long val[400000];
long long c[400000];
long long sig;
long long ans=-1;
int f[1<<18];
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%lld",&val[i]),sig+=val[i];
for(int i=1;i<=m;i++)
{
long long tmp;
scanf("%lld",&tmp);
c[i]=c[i-1]+tmp;
}
c[m+1]=1000000000000ll;
for(int i=0;i<(1<<n);i++)
{
long long sum=0;
for(int j=1;j<=n;j++)
if(1<<(j-1)&i) sum+=val[j];
for(int j=1;j<=n;j++)
if(1<<(j-1)&i)
{
int tmp=f[i-(1<<(j-1))];
int l=tmp+1,r=m+1;
while(l<r)
{
int mid=(l+r)>>1;
if(c[mid]-c[tmp]<=val[j]) l=mid+1;
else r=mid;
}
f[i]=max(f[i],l-1);
if(l-1==m&&ans<=sig-sum) ans=sig-sum;
}
}
printf("%lld",ans);
return 0;
}