POJ1276 Cash Machine 【解法一】

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.

解法二见【这里】。

二进制优化多重背包。

#include<cstdio>
#include<cstring>
#include<cmath>
bool dp[100010];
int vv[100010],v[15],num[15];
int main()
{
int i,j,k,m,n,p,q,x,y,z;
while (scanf("%d%d",&k,&n)==2)
{
memset(dp,0,sizeof(dp));
for (i=1;i<=n;i++)
scanf("%d%d",&num[i],&v[i]);
m=0;
for (i=1;i<=n;i++)
{
for (x=1;x<=num[i];x<<=1)
{
vv[++m]=x*v[i];
num[i]-=x;
}
if (num[i]) vv[++m]=num[i]*v[i];
}
dp[0]=1;
for (i=1;i<=m;i++)
for (j=k;j>=vv[i];j--)
dp[j]|=dp[j-vv[i]];
for (i=k;;i--)
if (dp[i])
{
printf("%d\n",i);
break;
}
}
}