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
一道简单的dp,思想就是记忆化。
代码如下:
/* ID: yizeng21 PROB: stamps LANG: C++ */ #include<stdio.h> #include<string.h> using namespace std; int dp[2000005]; int maxx; int max(int i,int j){ return i>j?i:j; } int min(int i,int j){ return i<j?i:j; } int a[10000]; int main(){ freopen("stamps.in","r",stdin); freopen("stamps.out","w",stdout); int m,n; scanf("%d%d",&m,&n); for(int i=1;i<=n;i++){ scanf("%d",&a[i]); maxx=max(maxx,a[i]); } memset(dp,10,sizeof(dp)); dp[0]=0; for(int j=1;j<=n;j++) for(int i=a[j];i<=maxx*m;i++){ dp[i]=min(dp[i],dp[i-a[j]]+1); } int tot=0; while(dp[tot]<=m)tot++; printf("%d\n",tot-1); }