374  Big Mod
Moderator: Board moderators
Hi!
Could anyone give test data with answers...
I have no idea why I get WA.
My code is:
#include <stdio.h>
unsigned long int b,p,m,res;
main()
{
while (scanf("%d %d %d",&b,&p,&m)==3)
{
res=1;
while (p>1) {
if (p&0x01==1)
res=((res%m)*(b%m))%m;
b=((b%m)*(b%m))%m;
p=p/2;
}
printf("%dn",((res%m)*(b%m))%m);
}
}
Thanks.
Could anyone give test data with answers...
I have no idea why I get WA.
My code is:
#include <stdio.h>
unsigned long int b,p,m,res;
main()
{
while (scanf("%d %d %d",&b,&p,&m)==3)
{
res=1;
while (p>1) {
if (p&0x01==1)
res=((res%m)*(b%m))%m;
b=((b%m)*(b%m))%m;
p=p/2;
}
printf("%dn",((res%m)*(b%m))%m);
}
}
Thanks.

 A great helper
 Posts: 284
 Joined: Thu Feb 28, 2002 2:00 am
 Location: Germany
 Contact:
What's the idea?
Hi, Pete!
Can u tell me what's the idea behind your code?
I can't figure out how to solve this problem... Did u used any MOD property?
Thanx in advance!!!
Jo
Can u tell me what's the idea behind your code?
I can't figure out how to solve this problem... Did u used any MOD property?
Thanx in advance!!!
Jo

 A great helper
 Posts: 284
 Joined: Thu Feb 28, 2002 2:00 am
 Location: Germany
 Contact:
Subeen: I used the same algorithm as pete did above. Can you tell us more about your algorithm, for example what it's name is? Otherwise it's hard to find what you mean...
jpfarias: pete basically just calculates it with the method of repeated squaring and uses mod everywhere to keep the intermediate results low.
Repeated squaring works like this: Imagine you want to calculate b^42. You could multiply 42 b's together, but that's slow. This is better:
b^42 = b^2 * b^8 * b^32
That's only 3 multiplications after you get the parts by
b^2 = b*b
b^4 = (b^2)^2
b^8 = (b^4)^2
b^16 = (b^8)^2
b^32 = (b^16)^2
So you start with b and then repeatedly multiply it with itself.
jpfarias: pete basically just calculates it with the method of repeated squaring and uses mod everywhere to keep the intermediate results low.
Repeated squaring works like this: Imagine you want to calculate b^42. You could multiply 42 b's together, but that's slow. This is better:
b^42 = b^2 * b^8 * b^32
That's only 3 multiplications after you get the parts by
b^2 = b*b
b^4 = (b^2)^2
b^8 = (b^4)^2
b^16 = (b^8)^2
b^32 = (b^16)^2
So you start with b and then repeatedly multiply it with itself.
374  I can't find the wrong
[c][/c]
Code: Select all
#include<stdio.h>
int main(void)
{
/* R = B^P mod M */
unsigned long B=0, P=0;
unsigned int M=0;
while(3==scanf("%d %d %d", &B, &P, &M))
{
int mod=1, i=0;
if(0==B%M) //best case
{
printf("0\n");
continue;
}
else if(B>M) //let B < M
B%=M;
while(1<P)
{
if(1==P%2)
{
mod*=B;
if(mod>M)
mod%=M;
}
P/=2;
B*=B;
if(B>=M) // let B < M
B%=M;
}
if(1!=mod)
mod=(mod*B)%M;
else
mod=B;
printf("%d\n", mod);
}
return 0;
}
Found Your mistake !!
Hi!
I have found your mistake...
It is rather stupid one ....
You have to look what will happen when P=0...
Than the answer should be 1 (except when the M=1 then it should be 0)
So try to change it and you will have AC..
Good Luck
I have found your mistake...
It is rather stupid one ....
You have to look what will happen when P=0...
Than the answer should be 1 (except when the M=1 then it should be 0)
So try to change it and you will have AC..
Good Luck
Stefan...
The following procedure computes a^c mod n as c is increased by doublings and incrementations from 0 to b.
Modular_Exponentiation ( a, b, n )
{
c = 0
d = 1
let {bk, bk1, …, b0} be the binary representation of b
for i = k downto 0
do c = 2 * c
d = (d * d) mod n
if bi = 1
then c = c + 1
d = (d * a) mod n
return d
}
/* The algorithm is collected from Cormen’s Intro. To Algo. */
Subeen.
The following procedure computes a^c mod n as c is increased by doublings and incrementations from 0 to b.
Modular_Exponentiation ( a, b, n )
{
c = 0
d = 1
let {bk, bk1, …, b0} be the binary representation of b
for i = k downto 0
do c = 2 * c
d = (d * d) mod n
if bi = 1
then c = c + 1
d = (d * a) mod n
return d
}
/* The algorithm is collected from Cormen’s Intro. To Algo. */
Subeen.
Last edited by Subeen on Wed Jul 24, 2002 7:46 am, edited 1 time in total.

 A great helper
 Posts: 284
 Joined: Thu Feb 28, 2002 2:00 am
 Location: Germany
 Contact:

 A great helper
 Posts: 284
 Joined: Thu Feb 28, 2002 2:00 am
 Location: Germany
 Contact:
Ok, I see. Sorry then... That unnecessary extra c stuff confused me. Also, you said sth like "The following procedure computes ac mod n [...]". This is a partly wrong (It's no multiplication, but an exponentiation) and partly counterintuitive (It computes (a^b)%n, not (a^c)%n) description. I know now how you meant it, but it's still a bit confusing. Sorry again...