Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 35022 | Accepted: 12706 |
Description
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
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
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
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
原題鏈接:http://poj.org/problem?id=1276
題意:有各種不同面值的貨幣,每種面值的貨幣有不同的數量,請找出利用這些貨幣可以湊成的最接近且小於等於給定的數字cash的金額.
樣例1:
要取735,取款機內有3種錢:4張125的,6張5的,3張350的.
735=3*125+2*5+1*350
動態規劃,多重揹包,可以二進制優化.
AC代碼:
#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;
}
參考代碼:
二進制優化: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