POJ 1276 - Cash Machine

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本文介绍了一种用于自动取款机(ATM)的现金分配算法，该算法旨在使用有限数量和种类的纸币来实现尽可能接近用户请求的现金数额。通过状态转移的方式，程序能够找出不超过目标金额的最大可分配现金值。

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

题意：给出一个目标金额和钞票种类，每种钞票给出面额和张数，问能够凑出最接近目标金额的值是多少。

控制在目标金额以下，每次取若干张钞票进行状态转移，记录最大值，最后输出最大值即可。

#include <cstdio>
#include <cstring>

bool ans[100005];
struct cash
{
    int counts;
    int val;
};
cash money[15];

int main()
{
    int goal, n;
    while (scanf("%d%d", &goal, &n) != EOF)
    {
        for (int i = 0; i < n; ++i)
            scanf("%d%d", &money[i].counts, &money[i].val);
        if (goal == 0 || n == 0)
        {
            printf("0\n");
            continue;
        }
        memset(ans, false, sizeof(ans));
        ans[0] = true;

        int maxn = 0;
        int tmp;
        for (int i = 0; i < n; ++i)
        {
            for (int j = maxn; j >= 0; --j)
            {
                if (ans[j] == true)
                {
                    for (int k = 0; k <= money[i].counts; ++k)
                    {
                        tmp = j + k * money[i].val;///状态转移
                        if (tmp > goal)
                            continue;
                        ans[tmp] = true;
                        if (tmp > maxn)
                            maxn = tmp;
                    }
                }
            }
        }
        printf("%d\n", maxn);
    }
    return 0;
}