| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 26172 | Accepted: 9238 |
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.
#include <iostream>
#include <cstdio>
using namespace std;
#define N 100003
int c[N],w[N];
int dp[N];
int n, v;
void ZeroOnePack(int cost,int weight) //01背包
{
for(int k=v; k>=cost; k--)
dp[k] = max(dp[k],dp[k-cost]+weight);
}
void CompeletPack(int cost,int weight) //完全背包
{
for(int k=cost; k<=v; k++)
dp[k] = max(dp[k],dp[k-cost]+weight);
}
void MultiplePack(int cost,int weight,int amount) //多重背包
{
if(cost*amount>=v)
{
CompeletPack(cost,weight);
return;
}
else
{
int k=;
while(k<amount)
{
ZeroOnePack(k*cost,k*weight);
amount = amount-k;
k=k*;
}
ZeroOnePack(amount*cost,amount*weight);
}
}
int main()
{
int i;
while(scanf("%d %d",&v,&n)!=EOF)
{
int totle = ;
int min = ;
for(i=; i<n; i++)
{
scanf("%d%d",&w[i],&c[i]); //每种金币的个数和金币面值
totle += w[i]*c[i];
if(min>c[i])
min = c[i];
}
if(totle<v) //如果金币的总量小于输入的v，直接将totle输出
{
printf("%d\n",totle);
continue;
}
if(v<min)//如果v小于金币的最小值，则直接输出0
{
printf("0\n");
continue;
}
memset(dp,,sizeof(dp));
for (i=; i<n; i++)
{
if(w[i]*c[i] == v)
{
dp[v] = v;
break;
}
else
MultiplePack(c[i],c[i],w[i]);//多重背包
} printf("%d\n",dp[v]);
}
return ;
}