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All I did was something like (within the pow function): where d is a double.

And in the end print the first three digits (it's easy in Java, at least it's good for something).

Darko wrote:All I did was something like (within the pow function):

Code: Select all

if(d > 1000000000) d /= 1000000000

where d is a double.

I m getting TLE when using the fast exponentiation method for Leading part also

I lied, this is what I did: And - how do you get TLE if you do logk operations?

Darko wrote:I lied, this is what I did:

Code: Select all

while (n > Integer.MAX_VALUE)
	n /= 1000000000;

And - how do you get TLE if you do logk operations?

No I do fast exponentiation now and geting WA

I got WA
Can anyone give a case where it fails down

#include<iostream>
#include<math.h>
#include<stdio.h>
using namespace std;
double bigMod1(double b,double p)
{
if (p==0)return 1.0;
long pp=(long)p;
if (pp % 2 == 0)
{
double r=bigMod1(b,p/2);
double j=(r*r);
if(j> 1000000000) j /= 1000000000;
return j;
}
else
{
double j=(b*bigMod1(b,p-1));
if(j> 1000000000) j /= 1000000000;
return j;
}
}
long bigMod(long b,long p,long m)
{
if (p==0)return 1;
if (p % 2 == 0)
{
long r=bigMod(b,p/2,m);
return (r*r)%m;
}
else
{
return ((b%m)*bigMod(b,p-1,m))%m;
}
}
int main(){
int n,m,k;
cin>>n;
while(n--){
cin>>m>>k;
long ttt;
ttt=bigMod(m,k,1000);
double mm=m,kk=k;
double lll=bigMod1(mm,kk);
while(lll<100){
lll*=10;
}
char buf1[30];

sprintf(buf1, "%.0Lf", lll);
cout<<buf1[0]<<buf1[1]<<buf1[2];


cout<<"...";
if(ttt<10){
cout<<"00"<<ttt<<endl;
}
else if(ttt<100){
cout<<"0"<<ttt<<endl;
}
else{
cout<<ttt<<endl;
}
}
return 0;
}

Cho wrote:

little joey wrote:I also calculated the first three digits using <math.h> and got accepted, but I still have some doubts about precision.
How can we be sure that a 100-million digit number that starts off "123999999999999999999999999..." is not printed as "124..." without doing bigint calculus?

I think the 4th to 15th digit can be considered as some sort of pseudo-random number, then the probability that all these digits being 9 is exponentially small.

Hmm, within the set of (2^32 - 1)*10^7 (approx. 4*10^16) possible input combinations, there will be an expected 40000 such cases, and an equal number of cases that get erroniously rounded down. I know of problems that test special cases with slimmer chances...

vinay pointed to me that my AC solution failed on cases with n = 10 and k = 100000 - it would give 999...000. Well, it was broken (at least in my and vinay's environment) for all k > 7 (I got either 10E...000 or 999...000).

I fixed it now, but the point is - I don't think they looked for nasty test cases. Or they omitted them on purpose - this problem ended up being the easiest problem in the set (judging by the number of AC during the contest).

Hi, I have submitted this problem several times but only managed to get T.L.E. I find the trailing digits by bigmod function and leading digits by fast exponentiation. Followings are the part of my code: and Any help is appreciated...

It worked for me, why don't you just make BIG a power of ten and then divide by it, instead of converting it back and forth?

BIG is actually power of 10 here, as you 've mentioned; 10^4 to be precise.10000l ... here. l is the suffix to denote long . Anyway, it is not working for me, I again got TLE after changing the code to divide by BIG.

Sorry, I should pay more attention - I used doubles for the base. And BIG 10^9.

But I am not sure why you are getting TLE. If you are dividing by BIG. If you are converting, what happens if you sprintf an 18-digit number into char[10]? I really don't know.

nymo wrote:Hi, I have submitted this problem several times but only managed to get T.L.E. I find the trailing digits by bigmod function and leading digits by fast exponentiation. Followings are the part of my code:

Code: Select all

FUNCTION USED TO FIND b^p mod m:
=========================
long long bigmod (long long b, long long p, long long m) 
{
	if (p == 0)
		return 1;
  
	else if (p % 2 == 0)
		return (bigmod (b, p / 2, m) * bigmod (b, p / 2, m)) % m; 
  
	else
		return ((b % m) * bigmod(b, p - 1, m)) % m;
}

if p is even you are computing bigmod b, p/2 twice instead of once in each debth of recursion. write something like
long long tmp = bigmod( b, p/2, m ) % m; tmp *= tmp; tmp %= m;
return tmp;

Hi!!!
I think there is a bug in this problem!
Here such input:
1
2 3

I think that answer should be : 008...008, but my accepted programm return 800...008
What do you think about it?

The statement says n^k will contain at least 6 digits, so don't worry.