USACO3.15stamps（dp）

对dp很无奈。。枚举所有可能达到的值 dp[i]表示到达i值所用最少的邮票 

/*
    ID: shangca2
    LANG: C++
    TASK: stamps
*/
#include <iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int p[55],dp[2000010];
bool o[2000010];
int main()
{
    freopen("stamps.in","r",stdin);
    freopen("stamps.out","w",stdout);
    int i,j,k,n;
    cin>>n>>k;
    for(i = 1 ; i <= k ; i++)
    {
        cin>>p[i];
        o[p[i]] = 1;
        dp[p[i]] = 1;
    }
    for(i = 1 ; ; i++)
    {
        if(o[i]&&dp[i]<n)
        {
            for(j = 1 ; j <= k ;j++)
            {
                o[i+p[j]] = 1;
                if(dp[i+p[j]]==0)
                dp[i+p[j]] = dp[i]+1;
                else
                dp[i+p[j]] = min(dp[i]+1,dp[i+p[j]]);
            }
        }
        else if(!o[i])
        {
            printf("%d\n",i-1);
            break;
        }
    }
    return 0;
}