mod operation on the > 10 bit

[sanjay123456]

Sir

i want that mod operation on gr8ter than 10 bit by following code of C

 long d,c,b,n;
 int i;
 d= 65073305
 c =7776;
n=162710186;
 b=1;
		 for(i=1;i<=d;i++)
		 {
			 b=(b*c);
			 b=b%n;
		  }

when loop is is run two time b= 60466176%162710 186 than b = 604666176

when loop is excuted three time than b =470184984576% 162710186
but its a wrong because b is long integer and its range is only 10 bit and our
b is 12 bit so mod operaion is also give wrong solution.......
How i solve this problem ?
plz anyone solve this problem


thanks

sanjay

[sivadhas2006]

Hi,

Try this...

unsigned long 
   b = 0;

I think the unsigned long variable is used to store the largest integer variable.

Regards,
M.Sivadhas.

[sanjay123456]

Sir ,

its not solved by unsigned long even if i tried long long that is hold 8 bytes

but its also print only 10 bit decimal data i want approxmately 20 bit data in decimal format
whay it so happened i dont understand
i am using madriva linux 2006 .

sanjay

[sivadhas2006]

Hi,

Can I know what is purpose of your program?
So there may be some other possible solution to solve the problem.

Regards,
M.Sivadhas.

[sanjay123456]

Sir ,

i want to implement RSA algorithms

sanjay

[sanjay123456]

Sir ,

whole code r following

#include<stdio.h>
#include<math.h>
long prime();

int gcd(long m);
long ptr(long m,int e);
main()
{

	 unsigned short int  p,q;
	 int i,e,M;
	 long n,m,d,c,b;
	 
	 long long mt;
		 
		 p=prime();
		 printf("p=%d\n",p);
		 q=prime();
		 printf("q=%d\n",q);
		 n=p*q;
		 printf("n=%ld\n",n);
		 m=(p-1)*(q-1);
		 printf("m=%ld\n",m);
		 e=gcd(m);
		 e=5;
		 printf("e=%d\n",e);
		 d=ptr(m,e);
		 printf("d=%ld\n",d);

		 M=6;
		 printf("M=%d\n",M);
		 b=1;
		 for(i=1;i<=e;i++)
		 {
			 b=b*M;
		 }
				 
		 c=b%n;
		 printf("b=%ld\n",b);
		 printf("cyper text=%d\n",c);	 
		 mt=1;
		 for(i=1;i<=d;i++)
		 {
			 mt=(mt*c);
			 mt=mt%n;
		 }

		 printf("plain text=%ld\n",mt);
	 

		 return 0;
}

long prime()
{
   long odd;
   odd = rand();
   return odd;
}
   
int gcd(long m)
{
	int i;
	for(i=2;i<m;i++)
	{
		if(m%i!=0)
		 return i;
	}
	
}	

long ptr(long m,int e)
{
	int i,v;
	long d;

	for(i=0;i<100;i++)
	{
		
	 v=(1+i*m)%e;
	 if(v==0)
	 {
		d=(1+i*m)/e;
		printf("i=%d\n",i);
		return d;

	 }
	}		 
	
}

sanjay