c语言65行代码哦 你们给几分？

#include <stdio.h>

int p1,p2,p3;
char in[1000];
int i;

int judge(){
if(in[i]=='\0')
return 1;
if((in[i]=='-'&&(i==0||in[i-1]>=in[i+1]||in[i-1]=='-'))||in[i]!='-'||(in[i]=='-'&&in[i-1]<60&&in[i+1]>60))
return 2;
if(in[i]=='-'&&in[i-1]<in[i+1])
return 3;
}

void function(){
int j,m;
if(p3==1){
for(j=in[i-1]+1;j<in[i+1];j++)
for(m=0;m<p2;m++){
if(p1==1)
printf("%c",j);
if(p1==2){
if(j>60)
printf("%c",j-32);
else
printf("%c",j);
}
if(p1==3)
printf("*");

}

}

if(p3==2){
for(j=in[i+1]-1;j>in[i-1];j--)
for(m=0;m<p2;m++){
if(p1==1)
printf("%c",j);
if(p1==2){
if(j>60)
printf("%c",j-32);
else
printf("%c",j);
}
if(p1==3)
printf("*");

}

}

}

int main(){
scanf("%d%d%d",&p1,&p2,&p3);
scanf("%s",&in);
for(i=0;i<1000;i++){
if(judge()==1)
break;
if(judge()==2)
printf("%c",in[i]);
if(judge()==3)
function();
}
return 0;
}

测试数据 #0: Accepted, time = 0 ms, mem = 432 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 432 KiB, score = 10

测试数据 #2: WrongAnswer, time = 0 ms, mem = 432 KiB, score = 0

测试数据 #3: Accepted, time = 0 ms, mem = 432 KiB, score = 10

测试数据 #4: WrongAnswer, time = 0 ms, mem = 432 KiB, score = 0

测试数据 #5: WrongAnswer, time = 0 ms, mem = 432 KiB, score = 0

测试数据 #6: Accepted, time = 0 ms, mem = 432 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 432 KiB, score = 10

测试数据 #8: Accepted, time = 15 ms, mem = 432 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 432 KiB, score = 10

求测试数据

var answer,s:ansistring;p1,p2,p3,i,n:longint;b:array[0..1000]of boolean;
function inabc(ll,rr:longint):boolean;
begin
if(s[ll]in['a'..'z'])and(s[rr]in['a'..'z'])then
exit(true);
end;
function pd(left,right:longint):boolean;
begin
if(s[left]in['a'..'z'])and(s[right]in['a'..'z'])then
exit(true);
if(s[left]in['0'..'9'])and(s[right]in['0'..'9'])then
exit(true);
exit(false);
end;
procedure writ(l,r:longint);
var ch:char;ii,f:longint;
begin
answer:='';
if not pd(l,r) then
begin
if b[l] then begin write(s[l]);b[l]:=false;end;
if b[l+1] then begin write(s[l+1]);b[l+1]:=false;end;
if b[r] then begin write(s[r]);b[r]:=false;end;
exit;
end;
if s[r]=succ(s[l]) then
begin
if b[l] then begin write(s[l]);b[l]:=false;end;
if b[l+1] then b[l+1]:=false;
if b[r] then begin write(s[r]);b[r]:=false;end;
exit;
end;
if s[r]=s[l] then
begin
if b[l] then begin write(s[l]);b[l]:=false;end;
if b[l+1] then begin write(s[l+1]);b[l+1]:=false;end;
if b[r] then begin write(s[r]);b[r]:=false;end;
exit;
end;
if s[l]>s[r] then
begin
if b[l] then begin write(s[l]);b[l]:=false;end;
if b[l+1] then begin write(s[l+1]);b[l+1]:=false;end;
if b[r] then begin write(s[r]);b[r]:=false;end;
exit;
end;
if b[l] then begin write(s[l]);b[l]:=false;end;
if p3=1 then
begin
answer:='';
for ch:=succ(s[l]) to pred(s[r]) do
for ii:=1 to p2 do answer:=answer+ch;
end
else if p3=2 then
begin
answer:='';
for ch:=pred(s[r]) downto succ(s[l]) do
for ii:=1 to p2 do answer:=answer+ch;
end;
if p1=2 then
if inabc(l,r) then
begin
f:=length(answer);
for ii:=1 to f do
answer[ii]:=chr(ord(answer[ii])-32);
end;
if p1=3 then
begin
f:=length(answer);answer:='';
for ii:=1 to f do answer:=answer+'*';
end;
write(answer);
if b[r] then begin write(s[r]);b[r]:=false;end;
if b[l+1] then b[l+1]:=false;
end;
begin
readln(p1,p2,p3);readln(s);n:=length(s);
fillchar(b,sizeof(b),true);
for i:=1 to n-2 do
if (s[i+1]='-') then writ(i,i+2)
else if b[i] then begin write(s[i]);b[i]:=false;end;
if b[n-1] then write(s[n-1]);
if b[n] then write(s[n]);
end.

#include<stdio.h>
#include<math.h>
#include<stdlib.h>

char s[100001];

main()
{
int p1,p2,p3;
long i,j,k,len;
char x;
scanf("%d %d %d",&p1,&p2,&p3);
len=0;
scanf("%c",&x);
x='0';
s[0]='A';
while (x!='\n')
{
scanf("%c",&x);
if (x=='\n') break;
len++;
s[len]=x;
}
for (i=1;i<=len;i++)
{
if (s[i]!='-') printf("%c",s[i]);
else if (((s[i-1]>='0')&&(s[i-1]<='9'))&&((s[i+1]>='0')&&(s[i+1]<='9')))
{
if (s[i-1]>=s[i+1]) printf("%c",s[i]);
else
{
if (p1!=3)
{
if (p3==1)
{
for (j=s[i-1]-'0'+1;j<=s[i+1]-'0'-1;j++)
for (k=1;k<=p2;k++) printf("%d",j);
}
else
{
for (j=s[i+1]-'0'-1;j>=s[i-1]-'0'+1;j--)
for (k=1;k<=p2;k++) printf("%d",j);
}
}
else
{
for (j=s[i-1]-'0'+1;j<=s[i+1]-'0'-1;j++)
for (k=1;k<=p2;k++) printf("*");
}
}
}
else if (((s[i-1]>='a')&&(s[i-1]<='z'))&&((s[i+1]>='a')&&(s[i+1]<='z')))
{
if (s[i-1]>=s[i+1]) printf("%c",s[i]);
else{
if (p1==1)
{
if (p3==1)
{
for (j=s[i-1]-'a'+1;j<=s[i+1]-'a'-1;j++)
for (k=1;k<=p2;k++) printf("%c",j+'a');
}
else
{
for (j=s[i+1]-'a'-1;j>=s[i-1]-'a'+1;j--)
for (k=1;k<=p2;k++) printf("%c",j+'a');
}
}
else if(p1==2)
{
if (p3==1)
{
for (j=s[i-1]-'a'+1;j<=s[i+1]-'a'-1;j++)
for (k=1;k<=p2;k++) printf("%c",j+'A');
}
else
{
for (j=s[i+1]-'a'-1;j>=s[i-1]-'a'+1;j--)
for (k=1;k<=p2;k++) printf("%c",j+'A');
}
}
else if (p1==3)
{
for (j=s[i-1]-'a'+1;j<=s[i+1]-'a'-1;j++)
for (k=1;k<=p2;k++) printf("*");
}
}
}
else printf("%c",s[i]);
}
printf("\n");
}

#include<stdio.h>
#include<string.h>
int p1,p2,p3;
char s[100001];

int main()
{
int l;
scanf("%d%d%d%s",&p1,&p2,&p3,s);
for(int i=0; i<strlen(s); i++)
if(s[i]=='-')
{
char c1,c2;
c1=s[i-1];
c2=s[i+1];
if(c1<c2)
{
if(c1>='0'&&c1<='9'&&c2>='0'&&c2<='9')
{
if(p1==3)
{
for(char j=c1+1; j<c2; j++)
for(int k=1; k<=p2; k++)
printf("*");
}
else
{
if(p3==1)
{
for(char j=c1+1; j<c2; j++)
for(int k=1; k<=p2; k++)
printf("%c",j);
}
else
{
for(char j=c2-1; j>c1; j--)
for(int k=1; k<=p2; k++)
printf("%c",j);
}
}
continue;
}
if(c1>='a'&&c1<='z'&&c2>='a'&&c2<='z')
{
switch(p1)
{
case 1:
{
if(p3==1)
{
for(char j=c1+1; j<c2; j++)
for(int k=1; k<=p2; k++)
printf("%c",j);
}
else
{
for(char j=c2-1; j>c1; j--)
for(int k=1; k<=p2; k++)
printf("%c",j);
}
break;
}

case 2:
{
if(p3==1)
{
for(char j=c1-31; j<c2-32; j++)
for(int k=1; k<=p2; k++)
printf("%c",j);
}
else
{
for(char j=c2-33; j>c1-32; j--)
for(int k=1; k<=p2; k++)
printf("%c",j);
}
break;
}
case 3:
{
for(char j=c1+1; j<c2; j++)
for(int k=1; k<=p2; k++)
printf("*");
break;
}
}
continue;
}
}
printf("-");
}
else printf("%c",s[i]);
return 0;
}

以后做这种题要小心点了，害我wa了三次

#include<string>
#include<iostream>
#include<cctype>
using namespace std;

string s;

void print(char ch,int n) {
for (int i=0;i<n;++i) cout<<ch;
}

bool pd(char a,char b) {
if (a>=b) return 1;
if (!(islower(a)&&islower(b))&&
!(isdigit(a)&&isdigit(b))) return 1;
return 0;
}

void solve(int p1,int p2,int p3) {
for (int i=0;i<s.size();++i) {
if (isalnum(s[i])||pd(s[i-1],s[i+1])||i==s.size()) cout<<s[i];
else {
if (p3==1) for (char ch=s[i-1]+1;ch<s[i+1];++ch) {
if (p1==1)
print(tolower(ch),p2);
if (p1==2)
print(toupper(ch),p2);
if (p1==3)
print('*',p2);
}
if (p3==2) for (char ch=s[i+1]-1;ch>s[i-1];--ch) {
if (p1==1)
print(tolower(ch),p2);
if (p1==2)
print(toupper(ch),p2);
if (p1==3)
print('*',p2);
}
}
}
}

int main() {
ios::sync_with_stdio(0);
int p1,p2,p3;
cin>>p1>>p2>>p3>>s;
solve(p1,p2,p3);
return 0;
}

编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 488 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 488 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 488 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 488 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 492 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 488 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 492 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 492 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 496 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 484 KiB, score = 10
Accepted, time = 0 ms, mem = 496 KiB, score = 100
#include<cstring>
#include<iostream>
#include<cstdio>
using namespace std;
char a,st,ed;
char ans[10000]={0};
int ansed=0;
int p1,p2,p3;
void work()
{
char temp[10000]={0};
memset(temp,0,sizeof(temp));
int c=0;

if(p1==1||(p1==2&&(st>='0'&&st<='9')))
for(int j=1;j<=ed-st-1;j++)
for(int i=1;i<=p2;i++)
temp[c++]=st+j;
if(p1==2&&st>'9')
for(int j=1;j<=ed-st-1;j++)
for(int i=1;i<=p2;i++)
temp[c++]=st+j+('A'-'a');
if(p1==3)
for(int j=1;j<=ed-st-1;j++)
for(int i=1;i<=p2;i++)
temp[c++]='*';
// ans[ansed++]=st;

if(p3==1)
for(int i=ansed;i<c+ansed;i++)

ans[i]=temp[i-ansed];
if(p3==2)
{int x=c+ansed-1;
for(int i=ansed;i<c+ansed;i++)

ans[x--]=temp[i-ansed];
}
ansed+=c;
}
int main()
{

scanf("%d%d%d",&p1,&p2,&p3);scanf("\n");
while(scanf("%c",&a))
{

if(a=='\n')
break;
if(a=='-')
{
scanf("%c",&ed);
if(ed=='\n')
{ans[ansed++]='-';
break;}
bool f=false;
if(st<='9'&&st>='0'&&ed<='9'&&ed>='0'&&ed>st)
f=true;
if(st<='z'&&st>='a'&&ed<='z'&&ed>='a'&&ed>st)
f=true;
if(f)
work(); if(!f)
ans[ansed++]='-';ans[ansed++]=ed;st=ed;
continue;
}ans[ansed++]=a;
st=a;
}printf("%s\n",ans);
return 0;
}

被一个回车折腾了好久。。。。

program p1379;
var
p1,p2,p3:longint;
a,b,s:char;
///////////////////////////////
procedure init;
begin

readln(p1,p2,p3);
end;
//////////////////////////////////
procedure main;
var i,j:longint;
begin
read(a,s,b);
while not(eof) do
begin
if s='-' then
begin
if (ord(a)>=ord(b)) or (a='-') or ((ord(a)<=57) and (ord(b)>=65)) then
begin
write(a);
a:=s;s:=b;read(b);
if eof then write(a,s,b);
end
else begin
write(a);
if p3=1 then
begin
if p1=1 then
begin
if ((ord(a)>=65) and (ord(b)<=90)) then
begin
for i:=ord(a)+33 to ord(b)+31 do
for j:=1 to p2 do write(chr(i));
end
else begin
for i:=ord(a)+1 to ord(b)-1 do
for j:=1 to p2 do write(chr(i));
end;
end;
if p1=2 then
begin
if (ord(a)>=97) and (ord(b)<=122) then
begin
for i:=ord(a)-31 to ord(b)-33 do
for j:=1 to p2 do write(chr(i));
end
else begin
for i:=ord(a)+1 to ord(b)-1 do
for j:=1 to p2 do write(chr(i));
end;
end;
if p1=3 then
begin
for i:=ord(a)+1 to ord(b)-1 do
for j:=1 to p2 do write('*');
end;

end;
if p3=2 then
begin

if p1=1 then
begin
if (ord(a)>=65) and (ord(b)<=90) then
begin
for i:=ord(b)+31 downto ord(a)+33 do
for j:=1 to p2 do write(chr(i));
end
else begin
for i:=ord(b)-1 downto ord(a)+1 do
for j:=1 to p2 do write(chr(i));
end;
end;
if p1=2 then
begin
if (ord(a)>=97) and (ord(b)<=122) then
begin
for i:=ord(b)-33 downto ord(a)-31 do
for j:=1 to p2 do write(chr(i));
end
else begin
for i:=ord(b)-1 downto ord(a)+1 do
for j:=1 to p2 do write(chr(i));
end;
end;
if p1=3 then
begin
for i:=ord(a)+1 to ord(b)-1 do
for j:=1 to p2 do write('*');
end;
end;
a:=b;read(s);
if eof then write(a,s)
else begin
read(b);
if eof then write(a,s,b);
end;
end;
end
else begin
write(a);
a:=s;s:=b;
read(b);
if eof then write(a,s,b);
end;
end;

end;
/////////////////////////////////
begin
init;
main;
end.

PASCAL...
这题太水...2min不会pascal照样过
program expand;
const a1=ord('0');a2=ord('9');b1=ord('a');b2=ord('z');

var i,j,k,l,m,n,p,q,r,t,p1,p2,p3:longint;
s,mid,mm:string;

procedure expanding(x:longint);
var g,t:longint;
begin
mid:='';
for g:=ord(s[x-1])+1 to ord(s[x+1])-1 do
for t:=1 to p2 do mid:=mid+chr(g);

if p1=2 then if ord(s[x-1]) in [b1..b2] then
for g:=1 to length(mid) do mid[g]:=chr(ord(mid[g])-32);

if p1=3 then for g:=1 to length(mid) do mid[g]:='*';

if p3=2 then
begin

mm:=mid;
for g:=1 to length(mid) do mid[g]:=mm[length(mm)-g+1];
end;

write(mid);
end;

function need(x:longint):boolean;
var g,t:longint;
begin
g:=ord(s[x-1]);t:=ord(s[x+1]);
if not(((g in [a1..a2])and(t in [a1..a2]))
or((g in [b1..b2])and(t in [b1..b2]))) then exit(false);
if ord(s[x-1])>=ord(s[x+1])-1 then exit(false);
exit(true);
end;

procedure work(x:longint);
begin
if s[x]<>'-' then
begin
write(s[x]);
exit;
end;

if x>1 then
begin
if need(x) then
begin
expanding(x);
exit;
end else
if ord(s[x-1])=ord(s[x+1])-1 then exit;
end;

write('-');
exit;
end;

begin
readln(p1,p2,p3);
readln(s);
s:=s+'$';
for i:=1 to length(s)-1 do work(i);
writeln;
end.

题目实在是太水了。。。可以看看这个18行的程序

#include

#include

#include

#include

int p1, p2, p3; std::string str, ans;

int main(void)

{

std::cin >> p1 >> p2 >> p3 >> str;

ans += *str.begin();

for (int i=1; i= str) || !((isdigit(str) == isdigit(str)) || (isalpha(str) == isalpha(str)))) {ans += str[i]; continue;}

for (char p=str+1; p

var

p1,p2,p3,l,k,i:integer;

s,ans:ansistring;

function work(p:longint):ansistring;

var left,right,i,j,k:longint;

begin

left:=ord(s[p-1]);

right:=ord(s[p+1]);

if left+1=right then exit('');

if left>=right then exit('-');

if (left=97) then exit('-');

if (left>=97) and (right=65) and (k=97) and (k

#include

#include

int p1,p2,p3;

void init()

{

scanf("%d%d%d",&p1,&p2,&p3);

}

char upc(char c)

{

return c-32;

}

void optr(char p)

{

int i;

if (p1==1)

{

for(i=1;i='a'&&p

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案错误...程序输出比正确答案长

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案错误...　├ 标准行输出

　├ 错误行输出

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Unaccepted 有效得分：80 有效耗时：0ms

function ad(h1,h2,h3,a1,a2:integer):ansistring;

var i,j:integer;

begin

case h1 of

1:begin

if h3=1 then for i:=a1+1 to a2-1 do

for j:=1 to h2 do ad:=concat(ad,chr(i))

else for i:=a1+1 to a2-1 do

for j:=1 to h2 do ad:=concat(chr(i),ad);

end;

2:begin

if a1>96 then begin

if h3=1 then for i:=a1+1 to a2-1 do

for j:=1 to h2 do ad:=concat(ad,chr(i-32))

else for i:=a1+1 to a2-1 do

for j:=1 to h2 do ad:=concat(chr(i-32),ad);

end

else begin

if h3=1 then for i:=a1+1 to a2-1 do

for j:=1 to h2 do ad:=concat(ad,chr(i))

else for i:=a1+1 to a2-1 do

for j:=1 to h2 do ad:=concat(chr(i),ad);

end;

end;

3:begin

for i:=a1+1 to a2-1 do

for j:=1 to h2 do ad:=concat(ad,'*');

end;

end;

end;

var p1,p2,p3,t,t1,t2:integer;

s,su,add:ansistring;

s1,s2:char;

begin

readln(p1,p2,p3);

readln(s);

t:=pos('-',s);

while t0 do begin

add:='';

su:=copy(s,0,t-1);

write(su);

s1:=s[t-1];

s2:=s[t+1];

t1:=ord(s1);

t2:=ord(s2);

if t1

选中查看源代码：

var p1,p2,p3:integer; s:string; ch:char; i,k,x,y,st,ed,j:integer;begin readln(p1,p2,p3); read(s); i:=1; while i1) and (i

program P1379(input,output);

const a1=['a'..'z'];

a2=['0'..'9'];

var s:array[1..999]of char;

j,k,g,h,z,p1,p2,p3,i:integer;

begin

assign(input,'P.in');

reset(input);

readln(p1,p2,p3);

z:=0;

while not eof do

begin

inc(z);

read(s[z]);

end;

close(input);

for i:=1 to z do

if s[i]'-' then write(s[i])

else if (i=1) or (i=z) then write('-')

else if (s='-')or(s='-') then write('-')

else

begin

g:=ord(s);

h:=ord(s);

if (g>=h)or((sin a1)and(sin a2))or((sin a2)and(s in a1)) then write('-')

else begin

if p1=3 then for j:=g+1 to h-1 do for k:=1 to p2 do write('*')

else if (h>48)and(h

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案正确... 0ms

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

第一次竟然忘记考虑‘--’的情况…… 终于秒杀 嘎嘎

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案错误...　├ 标准行输出

　├ 错误行输出

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Unaccepted 有效得分：90 有效耗时：0ms

我想要第六组数据，可以不？

#include

#include

#include

using namespace std;

string need;

int p1,p2,p3;

int temp1,temp2;

bool check(int num)

{

temp1=int(need[num-1]);

temp2=int(need[num+1]);

if (temp1> need;

for (i=0;i

本人一向不惮于把程序写的超长，为了正确 与 可读 ， 写的时候多复制粘贴了几下{}

var

i,j,k,n,t,p1,p2,p3:longint;

a:array[1..255]of char;

begin

readln(p1,p2,p3);

i:=1; while not eof do begin read(a[i]);inc(i) ;end; n:=i-1;

if (p1=1)and(p3=1) then

begin

i:=1;while a[i]='-' do begin write('-');i:=i+1;end;i:=i+1;

while i=ord(a))then begin write(a[i]);i:=i+2 end

else

begin

for j:=ord(a)+1 to ord(a)-1 do

for k:=1 to p2 do write(chr(j));

i:=i+2;

end;

end

else i:=i+1;

end;

if a[n-1]'-' then write(a[n-1]);writeln(a[n]);

end

else if (p1=1)and(p3=2)then

begin

i:=1;while a[i]='-' do begin write('-');i:=i+1;end;i:=i+1;

while i=ord(a))then begin write(a[i]);i:=i+2 end

else

begin

for j:=ord(a)-1 downto ord(a)+1 do

for k:=1 to p2 do write(chr(j));

i:=i+2;

end;

end

else i:=i+1;

end;

if a[n-1]'-' then write(a[n-1]);writeln(a[n]);

end

else if (p1=2)and(p3=1)then

begin

i:=1;while a[i]='-' do begin write('-');i:=i+1;end;i:=i+1;

while i=ord(a))then begin write(a[i]);i:=i+2 end

else

begin

if a

ansistring..

我的猥琐流……

procedure expand(x,y:char);

begin

x:=chr(ord(x)+1);

y:=chr(ord(y)-1);

if (p1=1) and (p3=1) then expand11(x,y)

else if (p1=1) and (p3=2) then expand12(x,y)

else if (p1=2) and (p3=1) and (x in ['a'..'z']) then expand21(x,y)

else if (p1=2) and (p3=2) and (x in ['a'..'z']) then expand22(x,y)

else if (p1=2) and (p3=1) and (x in ['0'..'9']) then expand11(x,y)

else if (p1=2) and (p3=2) and (x in ['0'..'9']) then expand12(x,y)

else if (p1=3) then expand3(x,y)

end;

然后写一个expandx，(CtrlC+CtrlV)*n......