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. 

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.

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.

题意：

给出总的钱币额V，给出n种币值和数目，问最接近V的的组合？

分析：

多重背包的模板题，多重背包问题是： 
有N种物品和一个容量为V的背包。第i种物品最多有num[i]件可用，每件费用是c[i]，价值是w[i]。求解将哪些物品装入背包可使这些物品的费用总和不超过背包容量，且价值总和最大。 
状态转移为：

F[i，v] = max{F[i−1,v−k∗Ci] + k∗Wi |0 ≤ k ≤ Mi}

对于本题，币值既是费用也是价值。

#include<iostream>

#include<algorithm>

#include<stack>

#include<cmath>

#include<string>

#include<cstring>

#include<cstdio>

#include<vector>

#include<deque>

using namespace std;

const int mx = ;

const int mxw = ;

deque<int>deqv, deq;

int w[mx], m[mx], dp[mxw];

int solve(int maxw, int n)

{

    memset(dp, , sizeof(dp));

    int i, k, j, val;

    for (i = ; i <= n; ++i)

        for (j = ; j < w[i]; j++)///枚举a=j%w[i]

        {

            while (!deqv.empty())//每枚举一次进行一次单调队列求值

            {

                deqv.pop_front();

                deq.pop_front();

            }

            for (k = ; k * w[i] + j <= maxw; k++)

            {

                val = dp[k * w[i] + j] - k * w[i];///划归为可重复使用的值

                while (!deqv.empty() && deqv.back() <= val)///保证deque的队首是最大的

                {

                    deqv.pop_back();

                    deq.pop_back();

                }

                deq.push_back(k);

                deqv.push_back(val);

                dp[k * w[i] + j] =deqv.front() + k * w[i];///从双端队列的头部取出最大值

                if (deq.front() == k - m[i])///这一头部已无法使用

                {

                    deq.pop_front();

                    deqv.pop_front();

                }

            }

        }

    return dp[maxw];

}

int main()

{

    int maxw, n, i;

    while (~scanf("%d%d", &maxw, &n))

    {

        for (i = ; i <= n; i++)

            scanf("%d%d", &m[i], &w[i]);

        printf("%d\n", solve(maxw, n));

    }

    return ;

}