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
这题还没有Noip1999那道邮票题难，那个还要搜索面值。
dp[i]表示构成i最少用几张
开始用n^2的dp过不完，结果发现是很简单的递推，反正从前面最少用的张数，推出现在需要的最少张数
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#define name "stamps"
using namespace std;
int k,n;
int s[55],dp[2000000];
int main()
{
freopen(name ".in","r",stdin);
freopen(name ".out","w",stdout);
cin>>k>>n;
int i,j;
for (i=1;i<=n;i++)
cin>>s[i];
i=1;
memset(dp,10,sizeof(dp));dp[0]=0;
while (true)
{
/*for (j=1;j<=n;j++)
if ((i/s[j])<dp[i]&&i%s[j]==0)
dp[i]=i/s[j];
for (j=1;j<=i/2;j++)
if (dp[j]+dp[i-j]<dp[i]) dp[i]=dp[j]+dp[i-j];*/
for (j=1;j<=n;j++)
if (i>=s[j]) dp[i]=min(dp[i],dp[i-s[j]]+1);
if (dp[i]>k) break;
i++;
}
cout<<i-1<<endl;
return 0;
}