POJ 1276 Cash Machine【多重背包】
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Cash Machine
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 35022 Accepted: 12706
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.
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.
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 350633 4 500 30 6 100 1 5 0 1735 00 3 10 100 10 50 10 10
Sample Output
73563000
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.
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.
Source
Southeastern Europe 2002
原题链接:http://poj.org/problem?id=1276
题意:有各种不同面值的货币,每种面值的货币有不同的数量,请找出利用这些货币可以凑成的最接近且小于等于给定的数字cash的金额.
样例1:
要取735,取款机内有3种钱:4张125的,6张5的,3张350的.
735=3*125+2*5+1*350
动态规划,多重背包,可以二进制优化.
AC代码:
#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>using namespace std;struct node{ int n,v;} a[20];int dp[100010];int main(){ int sum,n,i,j,k; //freopen("C:\\Documents and Settings\\Administrator\\桌面\\data.txt","r",stdin); while(~scanf("%d%d",&sum,&n)) { for(i = 1; i<=n; i++) scanf("%d%d",&a[i].n,&a[i].v); if(!sum||!n) { printf("0\n"); continue; } memset(dp,0,sizeof(dp)); dp[0] = 1; int MAX = 0,tem; for(i = 1; i<=n; i++) { for(j = MAX; j>=0; j--) { if(dp[j]) { for(k = 1; k<=a[i].n; k++) { tem = j+k*a[i].v; if(tem>sum) continue; dp[tem] = 1; if(tem>MAX) MAX = tem; } } } } printf("%d\n",MAX); } return 0;}参考代码:
二进制优化:http://www.cnblogs.com/zjbztianya/archive/2013/10/29/3394065.html
http://www.cnblogs.com/candy99/p/5796219.html
http://blog.csdn.net/zhou_yujia/article/details/51376157
http://blog.csdn.net/lyy289065406/article/details/6648102
http://www.cnblogs.com/372465774y/archive/2013/07/12/3187223.html
http://blog.csdn.net/zcmartin2014214283/article/details/51373299
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