有物品数量限制的背包问题

解决思路：

利用递归，1...N个物品，每种有Amount[i]个，那么从第N个物品开始选择，选0个、...、Amount[i]个第N个物品，依次计算价值，找出最大的，返回。

F(N) = max{ F(N-1, Lmt-n*Cost(n)) + n*Value(n), | n=0,..., Amount[i]}。

#ifndef HJD_SNACKBACK_H_
#define HJD_SNACKBACK_H_

int array_Weight[5]  = {10, 5, 3, 6, 2};
int array_Vacancy[5] = {5, 15, 8, 6, 5};
int array_Amount[5]  = {3, 3, 3, 5, 7};
int array_Price[5]   = {6, 10, 8, 6, 15};
int Weight_Lmt = 30;
int Vacancy_Lmt =30;

//递归去计算
int hjd_complete_snackback(int nPos, int nWeightLmt, int nVacancyLmt)
{
    int nAmount=0, nResult=0;
    if(nPos==0)
    {
        for(nAmount=0;
                (nAmount<=array_Amount[nPos])&&(nAmount*array_Weight[nPos]<=nWeightLmt)&&(nAmount*array_Vacancy[nPos]<=nVacancyLmt);
                nAmount++)
        {
            nResult=nAmount*array_Price[nPos];
        }
    }
    else
    {
        nAmount = 0;
        int ntmpResult = 0;
        while((nAmount<=array_Amount[nPos])&&(nAmount*array_Weight[nPos]<=nWeightLmt)&&(nAmount*array_Vacancy[nPos]<=nVacancyLmt))
        {
            ntmpResult = hjd_complete_snackback(nPos-1,
                                                nWeightLmt - nAmount*array_Weight[nPos],
                                                nVacancyLmt - nAmount*array_Weight[nPos]) + nAmount*array_Price[nPos];
            if (nResult


测试正确。