Backtracking Method 2nd Question-Stamp Issue

[Problem description]

There are m kinds of stamps with known denominations, each of which has n. Q: How many stamps can be combined consecutively with a total of no more than N stamps?

(1 <= m <= 225, 1 <= n <=, 1 <= stamp denomination <=)

Input: the value of the first line: N and M, separated by a space.

Row 2: A [1. m] (denomination), separated by a space.
Output: Maximum number of consecutive denominations

[Sample input]

4 31 2 4

[Sample output]

14

[Problem Analysis]

Programs I write

VaR A: array [0 .. 100] of integer; money, F: array [0 .. 255] of integer; n, m: integer; maxlong: integer; Procedure qsort (L, R: integer); // copy the VaR I from the FPC folder, j, X, Y: integer; begin I: = L; J: = r; X: = A [(L + r)> 1]; repeat while a [I] <X do Inc (I); While x <A [J] Do Dec (j); if not (I> J) then begin Y: = A [I]; A [I]: = A [J]; A [J]: = y; Inc (I); Dec (j); end; until I> J; if l <j then qsort (L, J); If I <r then qsort (I, R); end; Procedure Init; var I: integer; begin readln (n, m); for I: = 1 to M do read (A [I]); qsort (1, m); end; procedure search (k, n, x: integer); // K: the k-th known denomination, N: the maximum number of available known denominations, and X: The currently formed denomination value var I: integer; begin if (n = 0) or (k = 0) Then exit; for I: = n downto 0 do begin Inc (X, a [k] * I ); dec (n, I); Inc (money [x]); search (K-1, n, x); Inc (n, I); Dec (X, A [k] * I); end; Procedure count; var I, Max: integer; begin f [0]: = 0; max: = 0; for I: = 1 to 255 do if money [I]> 0 then begin f [I]: = f [I-1] + 1; if f [I]> MAX then Max: = f [I]; end else f [I]: = 0; writeln (max); end; begin Init; search (m, n, 0); count; end.

Standard procedure:

var  a:array  [1..100]  of  integer;  money:array  [1..2550]  of  integer;  total,n,m,i:integer;procedure  search(k,n,x:integer);var  i:integer;begin  if  (n=0)or(k=0)  then  exit;  for  i:=n  downto 0  do        begin           x:=x+a[k]*i;           n:=n-i;           money[x]:=money[x]+1;           search(k-1,n,x);           x:=x-a[k]*i;           n:=n+i;       endend;function maxlong:integer;var  j,total,max:integer;begin j:=n*a[m]; max:=0; repeat   while  (money[j]=0)and(j>0)  do      j:=j-1;   total:=0;   while  (money[j]>0)and(j>0)  do     begin total:=total+1;j:=j-1 end;   if  max<total then  max:=total   ; until  j<=0; maxlong:=maxend;begin  assign(input,‘word.in‘);  reset(input);  assign(output,‘word.out‘);  rewrite(output);  readln(m,n);  for  i:=1  to  m  do  read(a[i]);  total:=0;  search(m,n,0);  writeln(maxlong)end.

I really feel that this standard procedure is very good, and there are also dynamic planning practices for this question. Since I am learning the Backtracking Method, I will not introduce the DP Method for the moment.