USACO3.1.6--Stamps

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
题解：就是一个灰常简单的DP呀。和完全背包有点类似。这次的方程终于是我自己想出来的了！！！值得庆祝，哈哈。虽然方程很简单。。。虽然速度码出来了。但是提交上去就WA了。超内存了。。。开了个2000W的long数组，果断爆了。不过在自己的机器上居然没问题。。。难道是我的机器比较流弊吗？假设数组v为邮票的面值，f[i]表示组成面值为i邮票的最少数量。
方程：f[0]=0,f[i]=min(f[i],f[i-v[j]]+1)(1<=j<=k)

/*
ID:spcjv51
PROG:stamps
LANG:C
*/
#include<stdio.h>
#define MAXSN 2000000
int f[MAXSN];
int v[55];
int n,k;
long maxn;
int min(int a,int b)
{
    return a<b?a:b;
}
int main(void)
{
    freopen("stamps.in","r",stdin);
    freopen("stamps.out","w",stdout);
    long i,flag,j;
    scanf("%d%d",&n,&k);
    flag=1;
    maxn=0;
    for(i=1; i<=MAXSN; i++)
        f[i]=250;
    for(i=1; i<=k; i++)
    {
        scanf("%d",&v[i]);
        if(v[i]>maxn) maxn=v[i];
    }
    f[0]=0;
    maxn=maxn*n;
    for(i=1; i<=maxn; i++)
    {
        for(j=1; j<=k; j++)
            if(i-v[j]>=0&&f[i-v[j]]+1<=n)
                f[i]=min(f[i],f[i-v[j]]+1);
        if(f[i]==250) break;
    }
    printf("%ld\n",i-1);
    return 0;
}