回溯法实现邮票问题

/**//**
 *邮票问题
 *求最大连续区间，从邮资为1开始，步长为1
 *改进的子集合和函数 
 * @author by nowaywolf
*/

using System;
using System.Collections.Generic;
using System.Collections;
using System.Text;


namespace Stamp
...{
    class Program
    ...{
        static int[] price = new int[1000];
        static int m, n,r,t,current,max;
        static int[] vec = new int[1000];
        static bool flag = false;
        static void Main(string[] args)
        ...{
            price[0] = 0;
            Console.WriteLine("please enter the 面值总数:");
            n = Int32.Parse(Console.ReadLine());
            Console.WriteLine("please enter the 最大邮票张数:");
            m = Int32.Parse(Console.ReadLine());
            ArrayList StampPrice = new ArrayList();
            Console.WriteLine("please enter the stamp price:");
            
            //输入n种邮票面值
            for (int i=0; i <n; i++)
            ...{
                StampPrice.Add(Int32.Parse(Console.ReadLine()));
            }
            
            //计算集合中n*m张邮票的总邮资
            for (int i = 0; i < n; i++)
            ...{
                r+= (int)StampPrice[i] * m;
            }
            //构造一个n*m的集合
            for (int i = 0; i < n; i++)
            ...{
                for (int j = 0; j < m; j++)
                ...{
                    price[++t] = (int)StampPrice[i];
                }
            }
           
            //对于每一个邮票值，检查是否可得到
            for (int i = 1; i <= (int)StampPrice[n-1]*m; i++)
            ...{
                if (CheckNum(0, 1,r,i,0))
                ...{
                    Console.WriteLine("the number "+i+" is ok!");
                }
                if (flag&&(i==current+1))
                ...{
                    Console.WriteLine("["+"1"+","+i+"]");
                    current = i;//当前值可以到达，则保存以便求连续空间
                    max = i - 1;
                }
                else
                    if (flag)
                    ...{
                        Console.WriteLine("[]");
                    }
                flag = false;
            }
        }

        /**//// <summary>
        /// 检查邮资是否可达
        /// </summary>
        /// <param name="sum">用于累加的值</param>
        /// <param name="k">向量下标</param>
        /// <param name="r">当前邮资剩余值</param>
        /// <param name="M">要检查的邮资</param>
        /// <param name="count">用于控制邮票张数</param>
        /// <returns>如果所检查邮资可达，返回flag=true;否则返回flag=false</returns>
        public static bool CheckNum(int sum,int k,int r,int M,int count)
        ...{
            vec[k] = 1;
            if (count >= m)//邮票张数不能超过m
            ...{
                return flag;
            }
            if ((sum + price[k]) == M)//如果找到解
            ...{
                flag = true;
                Console.WriteLine();
                for (int i = 0; i <= k; i++)//输出该解是怎样构成的
                ...{
                    if (vec[i] != 0)
                    ...{
                        Console.Write(price[i] + "+");
                    }
                }
                return flag;
            }
            else
            ...{
                if (k<m*n&&sum + price[k] + price[k + 1] <= M)//k累加的次数不能超过m*n次
                ...{
                    CheckNum(sum + price[k], k + 1, r - price[k], M, count + 1);
                }
            }
            
            //没有找到解就回溯，如果找到了一个解就不再找其它解
            if (flag!=true&& sum + r - price[k] >= M && sum + price[k + 1] <= M)
            ...{
                vec[k] = 0;
                //count--;
                CheckNum(sum, k + 1, r - price[k],M,count);
                
            }
           return flag;
        }
    }
}