OJ Board

razibcse wrote:I can't understand how to calculate factorial of negative numbers...if u put n=0 in the formula (n-1)!=n!/n, u get
-1!=0!/0...how is this possible?

pls give some explanation of the process how negative even numbers get Underflow! and odd numbers get Overflow!...

Fact(0) = 1, But according Fact(n) = n * Fact(n - 1);
Fact(0) = 1 = 0 * Fact(-1)
So Fact (-1) is equal to "unlimited large" (Sorry, my English is poor )

We can also now Fact(-2) is "unlimited small"
(By Fact(-1) = -1 * Fact(-2) )

I think this is the specail case for this problem.

GoOd Luck

I think you don't consider one situation

that is if the input number is negative

and if the negative number is even, you should print "Underflow"

and if the negative number is odd, you should printf "Overflow"

if you want to know why?

look this rule

Factorial (n) = n*factorial (n-1).

you will find it.

But still I'm quite sure that negative factorial is undefined, has no answer (it's like division by zero). Negative factorial can't be solved in any ways known to mathematics. Besides there's a thread from Shahriar Manzoor telling us not ot bother with this prob. He also said the prob will be redefined.

According to http://mathworld.wolfram.com/Factorial.html...

... when n is a negative integer, in which case n! is equal to complex infinity.

Complex infinity is defined as:

An infinite number in the complex plane whose complex argument is unknown.

which as I understand it, is not exactly undefined as division by zero, but for our purposes in the "real plane world", I guess it is.

This is complete functional solution of this problem.
[c]
int main() {
long long p=1;
const long long o = 6227020800LL;
int i=0;

while(scanf("%i", &i) == 1){
if (i<0) {
if (-i%2==0) {
printf("Underflow!\n");
continue;
} else {
printf("Overflow!\n");
continue;
}
}
p=1;

for(; i>1; i--){
p*=i;
if (p>o) {
break;
}
}
if (p<10000) printf("Underflow!\n");
else if (p>o) printf("Overflow!\n");
else printf("%lli\n", p);
}
return 0;
}
[/c]

first i think its very easy.now i see its not so easy.i dont know what i have to do now.plz help me

/* @JUDGE_ID: xxxxxx 10323 C++ */


#include<string.h>
#include<math.h>
#include<stdio.h>
char t[1000];
static int sum,n;
long double sum1=0;
factorial(int k)

{
int i=0,c=0,sum,p;
p=strlen(t);
while(i<p)
{
sum=c+(t-48)*k;
t=(fmod(sum,10))+48;
i++;
c=sum/10;
}
while(c>0)
{
t[i++]=(c%10)+48;
c=c/10;
}
t='\0';
}
pr()
{
int i,j;
i=strlen(t);
if(i>10)
printf("Overflow!\n");
else if(i<4)
printf("Underflow!\n");
else
{
for(i=i-1;i>=0;i--)
{
sum1+=(t-48)*pow(10,i);
}
if(sum1>6227020800)
printf("Overflow!\n");
if(sum1<10000)
printf("Uuderflow!\n");
else
printf("%.0Lf\n",sum1);
}
sum1=0;
}

main()
{
int i;
while(scanf("%d",&n)!=-1)
{
strcpy(t,"1");
for(i=2;i<=n;i++)
{
factorial(i);
}
pr();
}
}

From a quick glance at your program, I can tell that it can't handle negative input. Perhaps you should see previous posts about this prob.
Good luck!!!

problem wrote:first i think its very easy.now i see its not so easy.i dont know what i have to do now.plz help me
...
if(sum1>6227020800)
printf("Overflow!\n");
if(sum1<10000)
printf("Uuderflow!\n"); =>missed something here??
...

And like Joseph say...
for negative input you don't need to find the result..
just check if it's odd or even.
The output should be Overflow if it's odd and Underflow if it's even.
Good Luck..
Regard,
Andre

Thank you,

I wasted about 5 submissions on this problem before reading this.
First I thought that if the number is negative , whether it is even I should output the factorial of its absolute value (|n|!) and otherwise - the negative factorial of its absolute value (-|n|!)
By the way , I have never heard of the factorial of a negative numer.
Thank you one more time

Thx Bug!
I was really frustrated coz factorial of negative number .. Thx


Aakash

can somebody tell me why it will compiler error?
[cpp]
#include <iostream>
using namespace std;

int main()
{
long long d;
int n;
int i;
bool f;
while(cin>>n)
{
f=false;
d=1;
for(i=1;i<=n;i++)
{
d*=i;
if(d>6227020800)
{
f=true;
break;
}
}
if(f)
cout<<"Overflow!"<<endl;
if(d<10000&&!f)
cout<<"Underflow!"<<endl;
if(d>=10000&&d<=6227020800&&!f)
cout<<d<<endl;
}

return 0;
}
[/cpp]

if(d>6227020800)

The problem could be over here. You are not allowed to access constant integer as big as this. You can use double to replace this.

place ll after each number, that is

[cpp]

if(d>6227020800ll)
{
f=true;
break;
}
}
if(f)
cout<<"Overflow!"<<endl;
if(d<10000ll&&!f)
cout<<"Underflow!"<<endl;
if(d>=10000ll&&d<=6227020800ll&&!f)
cout<<d<<endl;
[/cpp]

The ll indicates long long.

I made the correction and your code now gets WA.

Fixed how shamin said and you have to handle negative number with the funtion gama you have to see if is odd or even for each case you have 2 diferents anwers -1 overflow and -2 underflow
Hope it helps
Keep posting
P.S. Sorry for my english

why it gives me compile error

#include <stdio.h>

void main()
{
long double fact[6],f=5040,i,j;
long num;

for(i=8;i<=13;i++)
{
f=f*i;
fact[i-8]=f;
}

while(scanf("%ld",&num)==1)
{
if(num<8)
printf("Underflow!\n");
else if(num>13)
printf("Overflow!\n");

else printf("%.0Lf\n",fact[num-8]);
}
}