Cash Machine

Problem 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: <br> <br>cash N n1 D1 n2 D2 ... nN DN <br> <br>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. <br>  

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. <br>  

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
  
  
   多重背包模版題
  
  
   每行資料第一個是要求價值 共有幾件物品， 第一件物品的個數，第一件物品的價值.....
  
  
   要求找到能得到的距離要求最近的價值（不能比要求價值大）
  
  
          
#include<iostream>
#include<math.h>
#include<stdio.h>
using namespace std;

int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        int a[1000];
        int t=0;
        for(int i=0;i<m;i++)
        {
            int s1,v1;
            scanf("%d %d",&s1,&v1);
            for(int j=1;s1>0;j=j*2)
            {
                if(s1>=j)
                    s1=s1-j,a[t]=j*v1,t++;
                else
                    a[t]=s1*v1,s1=0,t++;
            }
        }
        int b[100010]={0};
        int ma=0;
        for(int i=0;i<t;i++)
        {
            for(int j=n;j>=a[i];j--)
                b[j]=max(b[j],b[j-a[i]]+a[i]),ma=max(ma,b[j]);
        }
        printf("%d\n",ma);
    }
    return 0;
}