(有限物品背包问题) poj 1276 Cash Machine

Cash Machine

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12957		Accepted: 4408

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

/*
题目大意：略；
题目解答：典型的有限背包求解问题。利用数学方法，讲有限物品背包换成0-1背包求解。
ps：详见牛人九解 http://blog.youkuaiyun.com/Hashmat/archive/2010/09/14/5883685.aspx
Source Code

Problem: 1276  User: wawadimu 
Memory: 564K  Time: 63MS 
Language: C++  Result: Accepted 

Source Code 
*/
#include<iostream>
using namespace std;

int f[100010];//f[v]重量为V的时候可行解？不行0：行1
int v[110];//第i中背包价值V[i]
int m;
int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d",&m)!=EOF)
    {
        int i,j,n,amut,w,k,cnt=0;
        scanf("%d",&n);
        while(n--)
        {
            scanf("%d%d",&amut,&w);
            k=1;
            while(amut-k>0)//散列成2^k
            {
                v[++cnt]=w*k;
                amut-=k;
                k*=2;
            }
            v[++cnt]=w*amut;

        }
        memset(f,0,sizeof(f));
        f[0]=1;//背包空状态
        for(i=1;i<=cnt;i++)
        for(j=m;j>=v[i];j--)//0-1背包 逆序计算
        f[j]=f[j] |f[j-v[i]];
        i=m;
        while(f[i]==0) i--;//输出第一个可行解
        printf("%d/n",i);
    }
    return 0;
}