OJ Board

Isn't it just calculate the last digit of M and last two digit of N and find the last sigit of the answer?
But I get Wrong Answer. Can anyone help me?
[pascal]Code is deleted[/pascal]

i got WA with my Pascal code.
and i could get AC by using scanf("%s%s") in C...

Iowai, what's the output of your accepted solution to this input :


0 238479283749827394234
1 2340982309420394802
2 2
2 5
3 3
3 74
4 982374982739487239847298374982374982734987
5 23849823498
6 2342
7 444444
8 3999949230402394
8 24
9 2345996959645456
2 0
209384032409803948209 4032094809238409238409
8932749872394729346628364 234972983749823749872394789234
0 0

0
1
4
2
7
9
4
5
6
1
4
6
1
1
9
6

hi lowai,

your words:

i got WA with my Pascal code.
and i could get AC by using scanf("%s%s") in C...

But i got AC usning scanf("%s %s") in C++.

But somebody can get AC with Pascal.
What's the problem of my program?

Accepted, thank you IOWAI, I was just messing up with my bignum library, I didn't notice that I wrote the digits backward in the vector. *(

but that's ok ... I've fixed that and got accepted. Nice problem. Thanks Shahriar Manzoor !

Why WA!

[pascal]
Const
modu: array [0..9] of longint = (1,1,4,4,2,1,1,4,4,2);
re: array [0..9,1..4] of byte =
((0,0,0,0), (1,0,0,0), (2,4,8,6), (3,9,7,1), (4,6,0,0),
(5,0,0,0), (6,0,0,0), (7,9,3,1), (8,4,2,6), (9,1,0,0));

Var
d,n,m: byte;

Procedure lee;
var
ch,cha: char;
begin
ch:= '0';
Repeat
cha:= ch;
read(input,ch);
until (ch=' ');
d:= ord(cha)-48;
ch:= '0';
Repeat
cha:= ch;
read(input,ch);
until (eoln(input));
n:= ord(ch)-48 + (ord(cha)-48 )*10;
end;

Begin
lee;
While (d<>0) or (n<>0) do
begin
m:= n mod modu[d];
if m=0 then m:= modu[d];
if n<>0 then writeln(output,re[d,m])
else writeln(output,1);
lee;
end;
End.
[/pascal]

Any help appreciated!

Have you tried the input I posted here?

Hello.
Pier. Your program reads the last digit of the first number and the last two af the second.
This is correct.
But your program will terminate with input (for example) :
556666496840 54655555555554600

So you have to get the lengths of digits too.
And I think there can be spaces before and after both digits. I think your program does not support it.

Good luck!

I don't know what makes my program redundant?

[c]
#include<stdio.h>

void main()
{ double m,n,i,temp;
int digit;
do{
scanf("%lf %lf",&m,&n);
digit=1;
if(!m && !m) break;
for(i=0;i<n;i++)
{ temp=(int)m%10;
digit*=(int)temp;
digit=digit%10;
}
printf("%d\n",digit);
}while(1);
}
[/c]

Eric try this:

100 100
0 100
0 0

I've been trying to solve this problem and i always get WA. I even tried to treat cases such spaces in the beginning and in the end of the lines, or numbers "0000000" or "000055543543", but still WA. Here is my code:

[pascal]
type bignr=
record
len:byte;
v:array[0..200] of byte;
end;
var n,m:bignr;
ch:char;
q,digit:integer;
function nulla(xyz:bignr):boolean;
var q:byte;
begin
for q:=1 to xyz.len do
if xyz.v[q]<>0
then
begin
nulla:=false;
exit;
end;
nulla:=true;
end;
begin
repeat
repeat
read(ch);
until ch in ['0'..'9'];
n.len:=1;
n.v[n.len]:=ord(ch)-48;
repeat
inc(n.len);
read(ch);
if ch in ['0'..'9']
then n.v[n.len]:=ord(ch)-48;
until not (ch in ['0'..'9']);
dec(n.len);
repeat
read(ch);
until ch in ['0'..'9'];
m.len:=1;
m.v[m.len]:=ord(ch)-48;
repeat
inc(m.len);
read(ch);
if ch in ['0'..'9']
then m.v[m.len]:=ord(ch)-48;
until not (ch in ['0'..'9']);
dec(m.len);
if
((m.len>1) or
(m.len=1) and
(m.v[1]<>0)) or

((n.len>1) or
(n.len=1) and
(n.v[1]<>0))
then
begin
if nulla(n) and nulla(m)
then digit:=0
else
begin
m.v[0]:=0;
if (n.v[n.len]=0)
then if nulla(m)
then digit:=1
else digit:=0
else
begin
digit:=1;
for q:=1 to m.v[m.len-1]*10+m.v[m.len]
do digit:=(digit*n.v[n.len]) mod 10;
end;
end;
writeln(digit);
end;
until (n.len=1) and (n.v[1]=0) and
(m.len=1) and (m.v[1]=0);
end.
[/pascal]

I desperately need help

Thanks, Overlord. I've got accepted.

[cpp]if(!m && !m) break;[/cpp]

You probably mean [cpp]if(!m && !n) break;[/cpp]

And I can tell you there are more.....
I wrote close to 150 lines to solve this problem.....takes 0.070 secs.