Solution to the problem of maximum continuous postage

Problem Description: suppose that the state issues n stamps of different denominations, and stipulates that only m stamps are allowed on each envelope. The continuous postage problem requires that for a given value of N and m, the optimal design of the face value of stamps is given, that is, the maximum continuous postage interval starting from postage 1 and increasing by 1 can be pasted on an envelope. For example, when n=5 and m=4, the maximum continuous postage range for five stamps with face value of (1,3,11,15,32) is 1-70.

Analysis:
For the continuous postage problem, n tuples x[1:n] are used to represent N Different Stamp denominations, and they are arranged from small to large. x[1]=1 is the only choice. At this time, the maximum continuous postage interval is [1:m].
The value range of x[2] is [2:m+1].
In general, if x[1:i-1] is selected, the maximum continuous postage interval is [1:r], then the value range of x[i] is [x[i-1]+1:r+1].

Algorithm design:

package com.bean.algorithmbasic;

public class StampsDemo2 {

    /*
    Continuous postage issues:
    Suppose that the state issues n stamps of different denominations, and stipulates that only m stamps are allowed on each envelope.
    The continuous postage problem requires that for a given value of n and m, the optimal design of stamp face value is given,
    The maximum continuous postage interval starting from postage 1 and increasing by 1 can be pasted on one envelope.

    For example:
    When n=2,m=3, if the face value is 1 and 4 respectively, then the postage can be obtained from 1 to 6 plus 8, 9 and 12. But 8, 9, 12 are already discontinuous sums.
    If the face value is 1,3, each postage value between 1 and 7 can be obtained, and 7 is the continuous maximum postage value.


    Resolver, recursive call, backtrace

    Input:
    2
    2 3
    5 4

    Output:
    7
    1 3

    70
    1 3 11 15 32
    */


    static final int NUM=10;
    static final int LEN=10000;
    static int[] x = new int[NUM];
    static int cnt = 0;//Current stamp type
    static int[] r = new int[NUM];//For face value results
    static int max = 0;//Maximum value

    static int[][] C = new int[NUM][LEN]; //Record search node status

    /*
     * Calculate the maximum continuous postage interval for the current face value of stamps
     * */
    public static int findMax(int knd, int lim) {
        int j = 1;
        while (C[cnt - 1][j]!=0) {
            if (j < x[cnt] || C[cnt - 1][j] <= C[cnt][j - x[cnt]] + 1)
                C[cnt][j] = C[cnt - 1][j];
            else
                C[cnt][j] = C[cnt][j - x[cnt]] + 1;
            j++;
        }

        while (true) {
            int tmp = Integer.MAX_VALUE;
            for (int i = 1; i <= cnt; i++) {
                if (tmp > C[cnt][j - x[i]] + 1)
                    tmp = C[cnt][j - x[i]] + 1;
            }
            if (tmp == Integer.MAX_VALUE || tmp > lim)
                break;
            else
                C[cnt][j] = tmp;
            j++;
        }
        C[cnt][j] = 0;
        return j - 1;
    }

    public static void dfs(int type,int limit) {
        if (cnt == type) {
            if (x[cnt] * limit < max)
                return;
            int tmp = findMax(type,limit);
            if (tmp > max) {
                max = tmp;//Update if it is larger than the upper limit of the maximum continuous postage interval recorded
                for (int i = 1; i <= type; i++) //Record new stamp face value combinations
                    r[i] = x[i];
            }
        }
        else {
            int tmp = findMax(type,limit);
            for (int i = tmp + 1; i >= x[cnt] + 1; i--) {
                x[++cnt] = i;//Add the possible face value to the current face value combination
                dfs(type,limit);
                cnt--;
            }
        }
    }





    public static void main(String args[]){
        int limit=3; //4 stamps allowed
        int type=2; //There are five denominations

        //From face value 1, there must be 1, otherwise many values cannot be constructed.
        x[1] = 1;
        //Because face value 1 is the first, cnt=1
        cnt = 1;
        dfs(type,limit);
        System.out.println("MAX = "+max);
        for (int i = 1; i <= type; i++) {
            System.out.println(r[i]+" ");
        }


    }




}

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