POJ1276 Cash Machine

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %lld & %llu

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.

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

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.

Source

Southeastern Europe 2002

多重背包 单调队列优化

 /*By SilverN*/
 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #define LL long long
 using namespace std;
 const int mxn=;
 int read(){
     int x=,f=;char ch=getchar();
     while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int lim,n;
 int m[mxn],v[mxn];
 bool f[];
 int q[];
 int main(){
     int i,j;
     while(scanf("%d%d",&lim,&n)!=EOF){
         memset(f,,sizeof f);
         for(i=;i<=n;++i){m[i]=read();v[i]=read();}
         f[]=;
         for(i=;i<=n;++i){
             if(m[i]==){
                 for(j=lim;j>=v[i];j--)
                     f[j]=f[j]||f[j-v[i]];
             }
             else{
                 for(int k=;k<v[i];++k){
                     if(k>lim)break;
                     int hd=,tl=;int smm=;
                     for(j=k;j<=lim;j+=v[i]){
                         q[++tl]=f[j];
                         smm+=f[j];
                         if(tl-hd>m[i])smm-=q[hd++];
                         f[j]=f[j]||smm;
                     }
                 }
             }
         }
         for(i=lim;i>=;--i)if(f[i]){
             printf("%d\n",i);break;
         }
     }
 }