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.

#include <iostream>

#include <stdio.h>

#include <cstring>

#include <algorithm>

using namespace std;

const int Max = ;

int cash, N, cnt[], d[];

int dp[Max];

void ZeroOnePack(int d)

{

    for (int i = cash; i >= d; i--)

        dp[i] = max(dp[i], dp[i-d]+d);

}

void CompletePack(int d)

{

    for (int i = d; i <= cash; i++)

        dp[i] = max(dp[i], dp[i-d]+d);

}

void MultiplePack(int m, int d)

{

    if (m*d >= cash) {

        CompletePack(d);

        return ;

    }

    int k = ;

    while (k < m) {

        ZeroOnePack(k*d);

        m -= k;

        k *= ;

    }

    ZeroOnePack(m*d);

}

int main()

{

    //freopen("1.txt", "r", stdin);

    while(cin >> cash) {

        cin >> N;

        for (int i = ; i <= N; i++)

            cin >> cnt[i] >> d[i];

        memset(dp, , sizeof(dp));

        for (int i = ; i <= N; i++)

            MultiplePack(cnt[i], d[i]);

        printf("%d\n", dp[cash]);

    }

    return ;

}

参考资料：背包九讲