easy

Why if there is no answer, we should print "5"? It must be included in problem description!!!

Pozhaluysta, kto-nibud', dayte mne hot' kakie testy k zadache 1070! A to vse testi, kotoriye u menya yest' moya programma prohodit, no na teste #1 dayot nepravil'niy otvet!
I obyasnite, pochemu mozhet byt' test:
"12.00 15.00
01.02 03.07
Answer: 0" ?

I have got WA at test 4,but I don't know why,I have debug my code for the whole afternoon,could you please give me some helpful tests?

01.01 21.59
04.23 11.22
Answer: 5

12.00 15.00
20.00 21.00
Answer: 1

23.58 00.43
22.27 03.10
Answer: 2

12.00 15.00
01.02 03.07
Answer: 0

01.01 10.59
04.23 04.22
Answer: 5

23.42 01.14
08.10 17.51
Answer: 4

"The difference between airport local times is not greater than 5 hours". This means that if you get an answer > 5, you should output 5 (it's stupid, i know). Got several WA6 before i realized it

Edited by author 05.09.2008 15:33

If you got Wrong answer    6 then
 if (answer>5) answer=5;
cout<<answer -- !!!

Most Important!!!

I have got AC when write it.

Example


01.01 21.59
04.23 11.22
Answer: 5

true 7 hours (21-7=7+7), but you must write 5!!!

good luck. :)

Double post.
Wrong!
please delete this post.

Edited by author 23.07.2010 02:19

Edited by author 23.07.2010 02:19

Where is Greenpeace? Oh,rebits,rebits...

Before sending solution, just test it on this examples.
1) 23.42 01.14
   08.10 17.51
Answer: 4
2) 01.01 10.59
   04.23 04.22
Answer: 5
3) 12.00 15.00
   01.02 03.07
Answer: 0
4) 23.58 00.43
   22.27 03.10
Answer: 2
5) 12.00 15.00
   20.00 21.00
Answer: 1
6) 01.01 21.59
   04.23 11.22
Answer: 5

;)

program a_local_time_1070;

{$APPTYPE CONSOLE}
{  1070 Pascal }
var to1,to2,tp1,tp2:real;
min1,min2:integer;
dd1,dd2,dd:integer;
arrpl:array[1..6] of boolean;
cou:integer;

function dif(t1,t2:real):integer;
var st1,st2,st3,st4:integer;
m:boolean;
diff:integer;
begin
m:=false;
diff:=0;
st1:=trunc(t1);
st2:=trunc((t1-st1)*100);
st3:=trunc(t2);
st4:=trunc((t2-st3)*100);
if (st1=st3) and (st2=st4) then m:=true;
while m=false do
begin
inc(diff);
inc(st2);
if st2=60 then
begin
inc(st1);
st2:=0;
end;
if st1=24 then
begin
st1:=0;
end;
if (st1=st3) and (st2=st4) then m:=true;
end;
dif:=diff;
end;

begin
readln(to1,tp1);
readln(to2,tp2);
dd1:=dif(to1,tp1);
dd2:=dif(to2,tp2);


if dd1<dd2 then
for cou:=1 to 6 do
begin
min1:=60*cou+dd1;
min2:=dd2-60*cou;
if (min1-min2>-11) and (min1-min2<11) then arrpl[cou]:=true;
end
else
for cou:=1 to 6 do
begin
min1:=60*cou+dd2;
min2:=dd1-60*cou;
if (min1-min2>-11) and (min1-min2<11) then arrpl[cou]:=true;
end;

for cou:=1 to 6 do
if arrpl[cou]=true then dd:=cou;

writeln(dd);

end.

{$APPTYPE CONSOLE}

var
 t1,t2,t3,t4: real;
 a1,a2,a3,a4: integer;

 k1,k2: integer;
 diff1, diff2: integer;

 i: integer;
begin
 Assign(input,'input.txt'); reset(input);
 Assign(output,'output.txt'); rewrite(output);

 readln(t1,t2);
 read(t3,t4);

 if t1 > t2 then t2 := t2 + 24;
 if t3 > t4 then t3 := t3 + 24;

 a1:=round(t1)*60;
 a2:=round(t2)*60;
 a3:=round(t3)*60;
 a4:=round(t4)*60;

 if  (t1-trunc(t1))>0.09 then  a1:=a1 + round((t1-trunc(t1))*100)
    else   a1:=a1  + round((t1-trunc(t1))*10);
  if  (t2-trunc(t2))>0.09 then  a2:=a2  + round((t1-trunc(t1))*100)
    else   a2:=a2  + round((t2-trunc(t2))*10);
  if  (t3-trunc(t3))>0.09 then  a3:=a3  + round((t3-trunc(t3))*100)
    else   a3:=a3  + round((t3-trunc(t3))*10);
  if  (t4-trunc(t4))>0.09 then  a4:=a4  + round((t4-trunc(t4))*100)
    else   a4:=a4  + round((t4-trunc(t4))*10);

 for i:=0 to 5  do
  begin
   diff1 := a1-i*60;
   diff2 := a4-i*60;
    k1:= abs(abs(diff1-a2));
    //writeln(k1 div 60,':',k1 mod 60, '   ',diff1,'-',a2,'  ', abs(diff1-a2));
    k2:= abs(abs(diff2-a3));
    //writeln(k1 div 60,':',k1 mod 60, '   ',diff2,'-',a3,'  ', abs(diff2-a3));

    //writeln(diff1 div 60,':', diff1 mod 60, '   ', t2:2:2);
    //writeln(t3:2:2,'   ',diff2 div 60,':', diff2 mod 60);


    if  abs(k1-k2)<=10 then writeln(i);
   end;
end.

Just use a simple math calculation, it's pretty easy..

Edited by author 22.08.2008 12:20

how can i get AC

   var
   t1, t2, t3, t4 : real;
   a1, a2, a3, a4, x : integer;

begin
   readLn(t1, t2);
   read(t4, t3);
   if t1 > t2 then t2 := t2 + 24;
   if t4 > t3 then t3 := t3 + 24;
   a1 := TRUNC(t1);
   a2 := TRUNC(t2);
   a3 := TRUNC(t3);
   a4 := TRUNC(t4);
   t1 := (t1-a1)*5/3 + a1;
   t2 := (t2-a2)*5/3 + a2;
   t3 := (t3-a3)*5/3 + a3;
   t4 := (t4-a4)*5/3 + a4;
   x := TRUNC(abs(t1-t2+t3-t4)/2);
   writeLn(x);
end.

var
 as,be,bs,ae:real;

procedure init;
 begin
  readln(as,be);
  readln(bs,ae);
 end;

procedure ik;
 var
  I,J:integer;
  flytime1,flytime2:real;
 begin
  for I:=-5 to 5 do
   begin
    if be-i<as then be:=be+24;
    flytime1:=(be-i)-as;
    if ae<bs-i then ae:=ae+24;
    flytime2:=ae-(bs-i);
    if abs(flytime1-flytime2)<1/6 then
     begin
      write(abs(I));
      halt;
     end;
   end;
 end;

begin
 init;
 ik;
end.

input:
23.42 01.14
08.10 17.51
my output is 9
The differ of the two airport is -4 hours.
So the time as the destination local time is below:
23.42-4=19.42   01.14=25.14
08.10           17.51-4=13.51
first duration is 25.14-19.42=5.32
second duration is 13.51-08.10=5.41
so the answer is 5.41-5.32=9
Is it right??
Thanks

I dont get it: why the answer for the test
01.01 21.59
04.23 11.22
is 5?

var
at,be,bs,ae:real;

procedure init;
begin
 readln(at,be);
 readln(bs,ae);
end;

procedure ik;
var
 I,J:integer;
 flytime1,flytime2:real;
begin
 for I:=-5 to 5 do
  begin
   if be-i<at then flytime1:=(be+24-i)-at else flytime1:=be-i-at;
   if ae<bs-i then flytime2:=ae+24-(bs-i) else flytime2:=ae-(bs-I);
   if abs(flytime1-flytime2)<1/6 then
    begin
     write(abs(I));
     halt;
    end;
  end;
end;

begin
 init;
 ik;
end.

Is it always a solution?
If not, what should I print?