10079 - Pizza Cutting

[Red Scorpion]

Just declare the variable as:
long long a;

sample:
int main() {
..long long a;
..a = 22345672948472475823;
.. printf ("%lld\n", a);
}

*for scanf and printf use the flag "%lld" *
Just that,

Good Luck!

[Sajid]

[c]long long a;
...
printf("%lld",...);[/c]

Is this both in the ANSI C ? plz, eXplain more.

[Jalal]

Its better to solve by a simple math law of High
school which u used to make
1+2+3+4+.....+n
just add 1 more to it.


Yes sajid that lld works with Turbo C
but not with borland C++

[anupam]

i think lld works in only linux and unix platform..
turbo c uses it but don't store actual amount of data that gcc store in linux..

[Dominik Michniewski]

anupam - you have right....

In C/C++ I use something like this (for 64 bit integer, I use MS VC++ as main compiler at home):

Code: Select all

#ifdef ONLINE_JUDGE /* parts for OJ system check */
#define LONG long long int
#define L_IN  "%lld"
#else /* Windows VC++ part */
#define LONG __int64
#define L_IN  "%I64d"
#endif

LONG a;

scanf(L_IN,&a);
printf(L_IN,a);

and it's work great. you could create more definitions i.e. for unsigned or other data type. BTW the same efect give you if you use typedef in C/C++ instead of #define as I

Dominik

[Eduard]

it gives WA help pleas
[pascal]program acm10079;
label 1;
var i,n:longint;
sum:comp;
begin
repeat
read(n);
if n<0 then goto 1;
sum:=n*n;
sum:=sum+n;
sum:=sum/2;
sum:=sum+1;
if n=0 then writeln(0) else writeln(round(sum));
until 1=2;
1:
end.
[/pascal]

[Russell John]

Your algorithm is inefficient.

[Eduard]

what you mean my algorithm is correct yes? then...?Or no...?

[osan]

any way u have to hold the result in long long or long double. coz it can be bigger than long range.

long double %Lf
long long %lld

result= (n-1)result+n
found the simple equation from that.
anyway if u can't then mail me. i will help u.

[midra]

First of all sorry for my bad english.
I think my formulas are right.
I I used this logic...
With every new line, the way that maximize the number of parts is that the new line pass over all the older lines, so with every new line you have n lines more, I mean, the line number five will add 5 more parts, for line 6 the pizza will have 6 more parts that when it has 5 lines, etc, etc...
From there I get the formulas but I get WA...

Here is my code:

[c]
CODE REMOVED
if somebody need help with this problem send me a PM
[/c]

If somebody knows why is wrong...please tell me.

[sumankar]

SIGSEGV is mainly due to invalid memory reference,
such as when you are trying to access invalid memory locs and/or
restricted mem-locs.

As for your program I can only suggest, provided your formula is correct
to change
[c]
while(scanf("%ld",&temp)!=EOF)
[/c]
to something like
[c]
while(scanf("%ld",&temp) == 1)
[/c]
that takes care of a missing EOF and other weird end-of-file stuff
hth
Suman

[midra]

Thank you 'Again' sumankar.
I have changed but I still got WA.... It never says SIGSEGV or an error like that, I always get WA...
I think maybe the type double is the problem, but I don't know...

[sumankar]

two things I'd like to add:
1)no need for double(i.e float-point arithmetic)

2)check your formula

There's a very thin line between helping and spoiling a problem,
for further help PM me/search the net.

hth
Suman

[sumankar]

two things I'd like to add:
1)no need for double(i.e float-point arithmetic)

2)check your formula

There's a very thin line between helping and spoiling a problem,
for further help PM me/search the net.

hth
Suman

[mhayter1]

On the problem cutting pizza. This works:

Code: Select all

cout<<((n*(n+1))/2)+1<<endl;

but this doesn't

Code: Select all

cout<<n*n*.5+n*.5+1<<endl;

you must change it to this:

Code: Select all

cout<<(long long)n*n*.5+n*.5+1<<endl;

does anyone know why this is?