USACO Section 3.1 Stamps（背包变形）

Stamps

Given a set of N stamp values (e.g., {1 cent, 3 cents}) and an upper limit K to the number of stamps that can fit on an envelope, calculate the largest unbroken list of postages from 1 cent to M cents that can be created.

For example, consider stamps whose values are limited to 1 cent and 3 cents; you can use at most 5 stamps. It's easy to see how to assemble postage of 1 through 5 cents (just use that many 1 cent stamps), and successive values aren't much harder:

• 6 = 3 + 3
• 7 = 3 + 3 + 1
• 8 = 3 + 3 + 1 + 1
• 9 = 3 + 3 + 3
• 10 = 3 + 3 + 3 + 1
• 11 = 3 + 3 + 3 + 1 + 1
• 12 = 3 + 3 + 3 + 3
• 13 = 3 + 3 + 3 + 3 + 1.

However, there is no way to make 14 cents of postage with 5 or fewer stamps of value 1 and 3 cents. Thus, for this set of two stamp values and a limit of K=5, the answer is M=13.

The most difficult test case for this problem has a time limit of 3 seconds.

PROGRAM NAME: stamps

INPUT FORMAT

Line 1:	Two integers K and N. K (1 <= K <= 200) is the total number of stamps that can be used. N (1 <= N <= 50) is the number of stamp values.
Lines 2..end:	N integers, 15 per line, listing all of the N stamp values, each of which will be at most 10000.

SAMPLE INPUT (file stamps.in)

5 2
1 3

OUTPUT FORMAT

Line 1:

One integer, the number of contiguous postage values starting at 1 cent that can be formed using no more than K stamps from the set.

SAMPLE OUTPUT (file stamps.out)

13
题意：假设你有 n 种不同面值的邮票，求使用小于等于 k 张邮票，第一个不能组成的值的前一个。
分析：d[i]表示组成面值为 i 的最少邮票数为 d[i]，于是 d[i]=min(d[i-a[j]]+1)；

/*
  ID: dizzy_l1
  LANG: C++
  TASK: stamps
*/
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
#define MAXN 51
#define MAXV 2000001
#define INF -0x7fffffff

using namespace std;

int d[MAXV],a[MAXN];

int main()
{
    freopen("stamps.in","r",stdin);
    freopen("stamps.out","w",stdout);
    int n,k,v,i,j;
    while(scanf("%d%d",&k,&n)==2)
    {
        v=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            v=max(v,a[i]);
        }
        v=v*k;
        for(i=1;i<=v;i++)d[i]=INF;
        d[0]=0;
        for(i=1;i<=v;i++)
        {
            int t=-INF;
            for(j=0;j<n;j++)
            {
                if(i-a[j]>=0)
                {
                    t=min(t,d[i-a[j]]+1);
                }
            }
            d[i]=t;
            if(d[i]<0||d[i]>k) break;
        }
        printf("%d\n",i-1);
    }
    return 0;
}