The minimum number of coins in exchange for change-dynamic programming problem 2

Enter m digits (positive number, must contain 1.) represents the face value of gold, and then enter n for the total amount of money exchanged, in order to obtain the minimum number of coins.

Dynamic Planning Issues 2

The basic idea of dynamic programming is to solve the problem of solving problems, solve the sub-problem, and save the solution of these sub-problems, if the solution of these sub-problems need to be solved later in solving the larger sub-problems, we can take out the calculated solutions and eliminate the repetition. The solution for saving sub-problems can be used as a form of fill-in, such as in arrays.

The code is as follows:

ImportJava.util.Scanner; Public classjin_bin_zhao_ling { Public Static voidMain (string[] args) {Scanner sc=NewScanner (system.in); System.out.print ("Enter the number of coin denominations:"); intM,n; M=Sc.nextint (); System.out.println ("Enter the face value of" +m+ "coins (must contain 1)"); inta[]=New int[M];  for(inti=0;i<m;++i) a[i]=Sc.nextint (); System.out.print ("Enter Total money N:"); N=Sc.nextint ();  Sc.close (); intb[]=Hui (a); N=Ge_shu (b,n); System.out.println ("The minimum number of coins is:" +n);}; Public Static intGe_shu (int[] A,intN) {    intLen=a.length;/*Note that the first element of the array is 0, to filter out*/    intf[]=New int[N+1]; f[0]=0;  for(intI=1;i<=n;++i)/*Core Algorithms*/    {        intj=1,temp=Java.lang.Integer.MAX_VALUE;  while(j<len&&i>=A[j]) {Temp=math.min (f[i-A[j]], temp); ++J; } F[i]=temp+1; }    returnf[n];}; Public Static int[] Hui (intA[]) {/*sorts the input data and rejects the duplicate numbers*/    intLen=1;/*return the length of the extra one*/     for(inti=0;i<a.length;++i) {intk=i;  for(intj=i+1;j<a.length;++j)if(a[j]<A[k]) k=J; if(k!=i) {inttemp=A[k]; A[K]=A[i]; A[i]=temp; }    }     for(inti=0;i<a.length;++i) {len++;  while((i+1) <a.length&&a[i]==a[i+1])/*Filter calls repeating elements*/++i; }    intb[]=New int[Len]; intk=1;/*intentionally increases the length by 1, so the preceding subscript 0 is empty. Easy to use after the subscript. */     for(inti=0;i<a.length;++i) {b[k]=A[i]; ++K;  while((i+1) <a.length&&a[i]==a[i+1])            ++i; }    returnb;};}

The minimum number of coins in exchange for change-dynamic programming problem 2