C++ :

#include <stdio.h>  
#include <string.h>  
#define MAX 10000000  
#define MIN -10000000  
int line[200],n,m,Min,Max,sum[200];  
int f[200][20];//f[i][j]=第1-i个数分成j份，结果最大值  
int g[200][20];//g[i][j]=第1-i个数分成j份，结果最小值  
int min(int a,int b)  
{  
    if(a>b)  
        return b;  
    return a;  
}  
int max(int a,int b)  
{  
    if(a>b)  
        return a;  
    return b;  
}  
void dp(int a[])  
{  
    int i,j,k;  
    for(i=1;i<=n;i++)  
        sum[i]=sum[i-1]+a[i];  
    for(i=0;i<=n;i++)  
        for(j=0;j<=m;j++)  
        {  
            f[i][j]=0;  
            g[i][j]=-1u>>1;  
        }  
    for(i=1;i<=n;i++)  
    {  
        f[i][1]=g[i][1]=(sum[i]%10+10)%10;  
    }  
    f[0][0]=1;  
    g[0][0]=1;  
    for(j=2;j<=m;j++)  
    {  
        for(i=j;i<=n;i++)  
        {  
            for(k=j-1;k<i;k++)  
            {  
                {  
                    f[i][j]=max(f[i][j],f[k][j-1]*(((sum[i]-sum[k])%10+10)%10));  
                    g[i][j]=min(g[i][j],g[k][j-1]*(((sum[i]-sum[k])%10+10)%10));  
                }  
            }  
        }  
    }  
    Max=max(Max,f[n][m]);  
    Min=min(Min,g[n][m]);  
}  
int main()  
{  
    //freopen("game.in","r",stdin);
    //freopen("game.out","w",stdout);
    int i,j,k;  
    Max=0;  
    Min=-1u>>1;  
    scanf("%d%d",&n,&m);  
    for(i=1;i<=n;i++)  
    {  
        scanf("%d",&line[i]);  
        line[i+n]=line[i];  
    }  
    for(i=0;i<n;i++)  
        dp(line+i);  
    printf("%d\n%d\n",Min,Max);  
    return 0;  
}  

Pascal :

uses math;
var
  f1,f2,c:array[0..1001,0..1001]of longint;
  a,b:array[0..1001]of longint;
  i,j,x,beg,n,k,ans1,ans2:longint;
begin
  readln(n,k);
  for i:=1 to n do
  begin
    readln(a[i]);
    a[i]:=a[i] mod 10+10;
    a[i+n]:=a[i];
  end;
	ans1:=maxlongint; ans2:=-maxlongint;
  for beg:=1 to n do
  begin
    for i:=1 to n do b[i]:=a[i+beg-1];
	  fillchar(c,sizeof(c),0);
	  for i:=1 to n do
	    for j:=i to n do
				c[i,j]:=((c[i,j-1]+b[j]) mod 10+10) mod 10;
    fillchar(f1,sizeof(f1),$7f);
		fillchar(f2,sizeof(f2),0);
		for i:=1 to n do
		begin
		  f1[i,1]:=c[1,i];
			f2[i,1]:=c[1,i];
		end;
		for j:=2 to k do
      for i:=j to n do
				for x:=j-1 to i-1 do
				  begin
					  f1[i,j]:=min(f1[x,j-1]*c[x+1,i],f1[i,j]);
						f2[i,j]:=max(f2[x,j-1]*c[x+1,i],f2[i,j]);
					end;
    ans1:=min(ans1,f1[n,k]); ans2:=max(ans2,f2[n,k]);
  end;
  writeln(ans1);
	writeln(ans2);
end.