Cash Machine
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 18125 |
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Accepted: 6307 |
Description
A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1,N, and for each denomination Dk the machine has a supply of nk bills. For example,
N=3, n1=10, D1=100, n2=4, D2=50, n3=5, D3=10
means the machine has a supply of 10 bills of @100 each, 4 bills of @50 each, and 5 bills of @10 each.
Call cash the requested amount of cash the machine should deliver and write a program that computes the maximum amount of cash less than or equal to cash that can be effectively delivered according to the available bill supply of the machine.
Notes:
@ is the symbol of the currency delivered by the machine. For instance, @ may stand for dollar, euro, pound etc.
Input
The program input is from standard input. Each data set in the input stands for a particular transaction and has the format:
cash N n1 D1 n2 D2 ... nN DN
where 0 <= cash <= 100000 is the amount of cash requested, 0 <=N <= 10 is the number of bill denominations and 0 <= nk <= 1000 is the number of available bills for the Dk denomination, 1 <= Dk <= 1000, k=1,N. White spaces can occur freely between the numbers in the input. The input data are correct.
Output
For each set of data the program prints the result to the standard output on a separate line as shown in the examples below.
Sample Input
735 3 4 125 6 5 3 350
633 4 500 30 6 100 1 5 0 1
735 0
0 3 10 100 10 50 10 10
Sample Output
735
630
0
0
Hint
The first data set designates a transaction where the amount of cash requested is @735. The machine contains 3 bill denominations: 4 bills of @125, 6 bills of @5, and 3 bills of @350. The machine can deliver the exact amount of requested cash.
In the second case the bill supply of the machine does not fit the exact amount of cash requested. The maximum cash that can be delivered is @630. Notice that there can be several possibilities to combine the bills in the machine for matching the delivered cash.
In the third case the machine is empty and no cash is delivered. In the fourth case the amount of cash requested is @0 and, therefore, the machine delivers no cash.
這個(gè)題描述不好，搞得我很亂，就是有一個(gè)cash大小的背包，有n中物品，每種物品有nk個(gè)，每個(gè)費(fèi)用dk，問包里最多能裝多少，
這是個(gè)多重背包，但是物品數(shù)特別多，有10*1000個(gè)最多，
剛開始沒注意，tle了，然后想到要用背包九講中講的二進(jìn)制拆分的思想
把每種物品分成1*w[i],2*w[i],4*w[i],……,2^k *w[i],(num[i]-2^(k+1)+1) *w[i] 這樣物品數(shù)就大大降低了，物品數(shù)變成了log（num）
因?yàn)槊總€(gè)數(shù)都可以用多個(gè)2的幾次方組成，
還有肯能的優(yōu)化是有的物品費(fèi)用超過了背包體積，直接舍去就可以了
1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4#define MAX 10000
5int v,n,w[MAX],num[20];
6int f[100100],ww;
7int i,j,k;
8int tot,t;
9int max(int a,int b)
10
{
11
if (a>b) return a;
12 else return b;
13
}
14int main()
15{
16 while (scanf("%d%d",&v,&n)!=EOF)
17
{
18 tot=0;
19 for (i=1; i<=n ; i++ )
20 {
21 scanf("%d%d",&num[i],&ww);
22 if (num[i]!=0)
23 {
24 tot++;
25 w[tot]=ww;
26 num[i]--;
27
}
28 t=2;
29 while (num[i]>=t)
30 {
31 num[i]-=t;
32 tot++;
33 w[tot]=t*ww;
34 t=t*2;
35 }
36 if (num[i]>0)
37 {
38 tot++;
39 w[tot]=num[i]*ww;
40 }
41 }
42 memset(f,0,sizeof(f));
43 for (i=1; i<=tot ; i++ )
44 for (j=v; j>=w[i]; j-- )
45 {
46 f[j]=max(f[j],f[j-w[i]]+w[i]);
47 }
48 {
49 }
50 printf("%d\n",f[v]);
51 }
52 return 0;
53}
54