| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 33444 | Accepted: 12106 |
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
一般解决方法：多重背包=01背包+完全背包
此处01处理时加上二进制思想，不然会超时
//524K 47MS C++ 742B
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std; int dp[];
int m[], d[];
int main(void)
{
int cash, n;
while(scanf("%d%d",&cash,&n)!=EOF)
{
memset(dp,,sizeof(dp));
for(int i=;i<n;i++){
scanf("%d%d",&m[i],&d[i]);
}
for(int i=;i<n;i++){
if(m[i]*d[i]>cash){
for(int j=d[i];j<=cash;j++)
dp[j] = max(dp[j], dp[j-d[i]] + d[i]);
}else{
int km=;
int k=m[i];
while(km<k){
int val = km*d[i];
for(int j=cash;j>=val;j--)
dp[j] = max(dp[j], dp[j-val] + val); k -= km;
km<<=;
} int val = k*d[i];
for(int j=cash;j>=val;j--)
dp[j] = max(dp[j], dp[j-val] + val);
}
}
printf("%d\n",dp[cash]);
}
return ;
}