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)
View Code
1 /* 2 ID:spcjv51 3 PROG:stamps 4 LANG:C 5 */ 6 #include<stdio.h> 7 #define MAXSN 2000000 8 int f[MAXSN]; 9 int v[55]; 10 int n,k; 11 long maxn; 12 int min(int a,int b) 13 { 14 return a<b?a:b; 15 } 16 int main(void) 17 { 18 freopen("stamps.in","r",stdin); 19 freopen("stamps.out","w",stdout); 20 long i,flag,j; 21 scanf("%d%d",&n,&k); 22 flag=1; 23 maxn=0; 24 for(i=1; i<=MAXSN; i++) 25 f[i]=250; 26 for(i=1; i<=k; i++) 27 { 28 scanf("%d",&v[i]); 29 if(v[i]>maxn) maxn=v[i]; 30 } 31 f[0]=0; 32 maxn=maxn*n; 33 for(i=1; i<=maxn; i++) 34 { 35 for(j=1; j<=k; j++) 36 if(i-v[j]>=0&&f[i-v[j]]+1<=n) 37 f[i]=min(f[i],f[i-v[j]]+1); 38 if(f[i]==250) break; 39 } 40 printf("%ld\n",i-1); 41 return 0; 42 }