C :

#include<stdio.h>
#include<string.h>
#define N 110
int main()
{
    char a[N];
    int i,n,aa[2],m,p1,p2,j,p3;
    scanf("%d %d %d",&p1,&p2,&p3);
    scanf("%s",a);
    for(i=0;a[i]!='\0';){
        if(a[i]=='-'){
            n=i;
            if(a[n-1]>='a' && a[n+1]<='z'){
                aa[0]=a[n-1]-'a';
                aa[1]=a[n+1]-'a';
                if(aa[0]>=aa[1]) printf("-");
                else if(aa[0]==aa[1]-1) printf("");
                else{
                    //printf("%c",aa[0]+'a');
                    if(p3==1){
                        for(m=aa[0]+1;m<aa[1];m++){
                            if(p1==1)
                                for(j=1;j<=p2;j++) printf("%c",m+'a');//小写字母；
                            else if(p1==2)
                                for(j=1;j<=p2;j++) printf("%c",m+'A');
                            else
                                for(j=1;j<=p2;j++) printf("*");
                        }
                    }
                    else{
                        for(m=aa[1]-1;m>=aa[0]+1;m--){
                            if(p1==1)
                                for(j=1;j<=p2;j++) printf("%c",m+'a');//小写字母；
                            else if(p1==2)
                                for(j=1;j<=p2;j++) printf("%c",m+'A');
                            else
                                for(j=1;j<=p2;j++) printf("*");
                        }
                    }
                    //printf("%c",aa[1]+'a');
                }
            }
            else if(a[n-1]>='0' && a[n+1]<='9'){
                aa[0]=a[n-1]-'0';
                aa[1]=a[n+1]-'0';
                if(aa[0]>=aa[1]) printf("-");
                else if(aa[0]==aa[1]-1) printf("");
                else{
                    //printf("%c",aa[0]+'a');
                    if(p3==1){
                        for(m=aa[0]+1;m<aa[1];m++){
                            if(p1==1 || p1==2)
                                for(j=1;j<=p2;j++) printf("%c",m+'0');
                            else
                                for(j=1;j<=p2;j++) printf("*");
                        }
                    }
                    else{
                        for(m=aa[1]-1;m>=aa[0]+1;m--){
                            if(p1==1 || p1==2)
                                for(j=1;j<=p2;j++) printf("%c",m+'0');
                            else
                                for(j=1;j<=p2;j++) printf("*");
                        }
                    }
                    //printf("%c",aa[1]+'a');
                }
            }
            else{
                printf("%c",a[i]);
            }
            i++;
        }
        else{
            printf("%c",a[i]);
            i++;
        }
    }
    return 0;
}

C++ :

//先存再输出是很愚蠢的方法！！
//一边改一边输出！！ 
#include <iostream> 
#include <cstdio>
#include <cstring>
using namespace std;
int p1,p2,p3;
int flag;
char t,i,flag2;
	
char change(char c){					//p1=2填充大写字母 
	return c-32;
}

void ptkz(char p){
	int i;
	if (p1==1){
		for(i=1;i<=p2;i++){
			cout<<p;				//直接输出小写字母 
		}
	}
	if (p1==2&&p>='a'&&p<='z'){
		for(i=1;i<=p2;i++){
			cout<<change(p);		//输出大写字母 
		}
	}
	if(p1==2&&p>='0'&&p<='9'){
		for(i=1;i<=p2;i++){
			cout<<p;				//填充数字串 
		}
	}
}
void kzxh(){
	int i;
	for(i=1;i<=p2;i++){
		cout<<"*";					//填充星号串 
	}
}

int judge(char a,char b){
	if(a>='a'&&a<='z'&&b>='a'&&b<='z')
	if(a<b) return 1;
	if(a>='0'&&a<='9'&&b>='0'&&b<='9')
	if(a<b) return 1;
	return 0;
}

int main(){
	cin>>p1>>p2>>p3;
	flag2='#';		//作为一个标志使用 
	while(cin>>t){
		if(t!='-')cout<<t;		//非 '-' 则直接输出即可 
		else{
			if(cin>>t){
				flag=judge(flag2,t);
				if(flag){
					if(p3==1){
						for(i=flag2+1;i<=t-1;i++){
							if(p1!=3)ptkz(i);else kzxh();
						}
					}
					else{
						for(i=t-1;i>=flag2+1;i--){
							if(p1!=3)ptkz(i);else kzxh();
						}
					}
					cout<<t;
				}
				else cout<<"-"<<t;
			}
			else cout<<"-";
		}
		flag2=t;
	}
	cout<<endl;
	return 0;
}

Pascal :

var
  p1,p2,p3,i:longint;
  from,en,ch:char;
  ans:ansistring;
  s:string;

function operate(ch:char):string;
  var rt:string;
    i:longint;
  begin
    if (ch in ['a'..'z'])and(p1=2)
      then ch:=chr(ord(ch)-32);
    if p1=3 then ch:='*';
    rt:='';
    for i:=1 to p2 do rt:=rt+ch;
    exit(rt);
  end;

begin
  readln(p1,p2,p3);
  readln(s);
  ans:=''; i:=1;
  while i<=length(s) do
    begin if s[i]='-' then
      begin if (i>1)and(i<length(s))and(((s[i-1] in['0'..'9'])and(s[i+1] in ['0'..'9'])and(s[i-1]<s[i+1]))
               or ((s[i-1] in ['a'..'z'])and(s[i+1] in ['a'..'z'])and(s[i-1]<s[i+1])))
              then begin
                     from:=chr(ord(s[i-1])+1);
                     en:=chr(ord(s[i+1])-1);
                     if p3=1
                  then for ch:=from to en do
                         ans:=ans+operate(ch)
                  else for ch:=en downto from do
                         ans:=ans+operate(ch);
                   end
              else ans:=ans+'-';
       end
       else ans:=ans+s[i];
    inc(i);
  end;
  writeln(ans);
end.