18 条题解
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0
fenghao LV 10 @ 2009-10-24 00:59:17
冒着被封号的危险，我来晒两大段程序吧：
第一段是这题的递归写法：
type rec=record
unknow:string;
value:longint;
end;
var cover:array[1..50] of rec;
s,ts:string;
n,i,k,total,c:longint;
function count(c:char; t:longint):longint;
begin
t:=t*10;
if c='+' then exit(1+t) else
if c='-' then exit(1+t) else
if c='*' then exit(3+t) else
if c='/' then exit(3+t);
end;
function find(s:string):longint;
var i:longint;
begin
while s[1]='0' do delete(s,1,1);
for i:=1 to total do
if cover[i].unknow=s then exit(cover[i].value);
end;
procedure block(var s:string);
var i,j,len:longint; flag:boolean;
begin
len:=length(s); j:=0;
if (s[1]='(')and(s[len]=')') then
begin
for i:=1 to len do
begin
if s[i]='(' then inc(j) else
if s[i]=')' then dec(j);
if (j=0)and(ilen) then exit;
end;
delete(s,1,1);
delete(s,length(s),1);
end;
end;
function cal(s:string):longint;
var i,len,c,t,delta,min,cut,temp:longint;
begin
block(s);
if s[1]'(' then s:='0'+s;
len:=length(s);
temp:=0;
val(s,temp,c);
if c0 then
begin
min:=maxint; cut:=0; delta:=0;
for i:=1 to len do
begin
if s[i]='(' then inc(delta) else
if s[i]=')' then dec(delta) else
if s[i] in ['+','-','*','/'] then
begin
t:=count(s[i],delta);
if t='0' do
begin
t:=t+st[a];
inc(a);
end;
for i:=1 to n do
if v[i].s=t then exit(v[i].c);
val(t,find);
end;
function cal(a:char):longint;
begin
case a of
'+': exit(1);
'-': exit(2);
'*': exit(3);
'/': exit(4);
'(': exit(5);
')': exit(6);
'#': exit(7);
end;
end;
begin
readln(t);
for m:=1 to t do
begin
readln(st);
readln(n);
for i:=1 to n do
begin
readln(v[i].s);
k:=pos('=',v[i].s);
val(copy(v[i].s,k+1,length(v[i].s)-k),v[i].c);
delete(v[i].s,k-1,length(v[i].s)-k+2);
end;
fillchar(num,sizeof(num),0);
fillchar(opt,sizeof(opt),#0);
st:=st+'#';
i:=1;
while i='0' then
begin
inc(p2);
num[p2]:=find(i);
end else
begin
case f[cal(opt[p1]),cal(now)] of
'>':
begin
num[p2-1]:=count(num[p2-1],num[p2],opt[p1]);
dec(p2); dec(p1);
end;
' -
0
lijian3256 LV 10 @ 2009-07-30 16:24:42
错把字符串当字符，苦调1小时终发现....
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0
pop542677493 LV 3 @ 2009-01-12 10:40:41
const
xiaoxie=['a'..'z'];
fuhao=['-','=','+'];
//>
function panduan(st:string):boolean;
var
i:longint;
begin
for i:=1 to length(st) do
if st[i] in xiaoxie
then exit(true);
exit(false);
end;
//>
procedure init;
begin
readln(a);
step1:=0;
step2:=1;
d:=pos('=',a);
end;
//
begin
init;
work;
write(ans);
end. -
0
181818181818 LV 10 @ 2008-10-23 19:06:59
实在是好事多磨，多少天了，我终于把这题干掉了！
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0
样例中应该是英文的 ':' 而不是中文的 '：'
然后正常做即可，处理负数的时候我是在负数前面统一加0处理
fenghao
LV 10
@ 2009-10-24 00:59:17
- 1
