10515 - Powers Et Al.

[osan]

Dear bayzid

i think it would be tuff for you to handle last two digit or or last digit. after getting input as a string.

i hope u know about the function strlen.
try to use this

hope you understand whatever i wanna to say.

if you can't mail me then.

[seadoo]

2 digits is not enough and 3 digits can do .

Nope, one digit of m and two digits of n are enough.

[Master]

Yes, I agree with seadoo.
one digit of m and two digits of n are enough.

M H Rasel
CUET Old Sailor

[Tahseen Mohammad]

Think about what happens if you calculate m^n. Think about the
manual multiplication steps.

You multiply every digit of m with n. There are some carry propagation
and others. But for the last digit there is no carry or other things. Try
simulating something like 15^15. See how the last digit remains 5^15.

Now we solve the problem of m being too big. But again n is also too big.
Now look at the results of 0^i through 9^i (only last digit). You will see
there is some repeating pattern. To cover all repeating patter with
modulus calculation it is necessary you take the last two digit of n and
then use the repeating pattern to get to solution.

So finally you have for any m, n you can have the result in constant time.

Hope it makes things a little clear.

[snake]

Hy! I'm getting TLE with this code. Can anyone explain why?

[cpp]
#include <iostream>
#include <math.h>
#include <stdio.h>
using namespace std;

/*
2 2
2 5
0 0

*/

int main()
{
int m,n;
while(cin>>m>>n)
{
if(!m && !n)
return 0;
if(m==1){
cout<<"1"<<endl;
continue ;
}
if(m==10 || m==0){
cout<<"0"<<endl;
continue;
}
int c=0;
int dig=1;
while(c++<n)
{
dig*=m;
if(dig>=10)
dig = dig-floor((double)dig/10)*10;

}
cout << dig << endl;

}
return 0;
}


[/cpp]

[Master]

Another word here i want to add that a big integer is divisable by 4, only its last two digit is divisable by 4. I mean a big integer such as
.......XX
is divisable by 4, if the number
XX is divisable by 4

I think it will help more

If you want more help then please visit the site
http://www.acmbeginner.tk

or

http://www.geocities.com/acmbeganer/Acm ... Tricks.htm

M H Rasel
CUET Old Sailor

[sohel]

Problem Description :

Each line contains two integers m and n (Less than 10^101).

From your source code it looks like you are reading the input as integers. I don't think
int data type can handle such a large number

You must read the numbers as string and then process.

Hope it helps.

[shamim]

Here is the Input Description

The input file contains less than 100000 lines. Each line contains two integers m and n (Less than 10^101). Input is terminated by a line containing two zeroes. This line should not be processed.

You are using integer variable for m and n, this is obviously wrong.

[snake]

Oh, I misunderstood the question then

Thanks for help. I'll gonna fix that.

[Rich]

I have tried lots of test data I could think about.

And I have read the posts about 10515,

but I still can't figure out what's the problem with my code......

Could someone PLS help me ?

thanks a lot......


[cpp]#include <stdio.h>
#include <string.h>
#include <math.h>

int Rich_Mod(char *b,int cycle)
{
int t;
if(cycle==1)
return 1;
if(cycle==2)
return (b[strlen(b)-1]-'0')%2+2;
if(cycle==4)
{
if(strlen(b)>1)
t = 10*b[strlen(b)-2]+b[strlen(b)-1];
else
t = b[strlen(b)-1];
return t%4+4;
}
}
void main()
{
char a[2000],b[2000];

int Base[10]={0,1,4,4,2,1,1,4,4,2};
int cycle,c;
int mode=0;

while(scanf("%s %s",&a,&b)==2)
{
if(strlen(a)==1 && a[0]=='0' && strlen(b)==1 && b[0]=='0')
break;
if(strlen(b)==1 && b[0]=='0')
printf("1");
else if(strlen(a)==1 && a[0]=='0')
printf("0");
else
{
cycle = a[strlen(a)-1]-'0';
c = Rich_Mod(b,Base[cycle]);
cycle = pow(cycle,c);
cycle%=10;
printf("%d",cycle);
}
printf("\n");

}
}[/cpp]

[osan]

GET THE INPUTS FROM FILE & CHECK THE OUTPUT WITH YOUR OUTPUT
INPUT

8 0
1 1
1 2
1 3
1 4
1 5
1 6
1 7
1 8
2 1
2 2
2 3
2 4
2 5
2 6
2 7
2 8
3 1
3 2
3 3
3 4
3 5
3 6
3 7
3 8
4 1
4 2
4 3
4 4
4 5
4 6
4 7
4 8
5 1
5 2
5 3
5 4
5 5
5 6
5 7
5 8
6 1
6 2
6 3
6 4
6 5
6 6
6 7
6 8
7 1
7 2
7 3
7 4
7 5
7 6
7 7
7 8
8 1
8 2
8 3
8 4
8 5
8 6
8 7
8 8
0 0

OUTPUT

1
1
1
1
1
1
1
1
1
2
4
8
6
2
4
8
6
3
9
7
1
3
9
7
1
4
6
4
6
4
6
4
6
5
5
5
5
5
5
5
5
6
6
6
6
6
6
6
6
7
9
3
1
7
9
3
1
8
4
2
6
8
4
2
6

[Rich]

^^...

thanks , I got accept.......

I have checked the input and the output,
but there is no problem (I mean the output is the same as yours...)


at the same time , I send my code again........

Unbelievably, I got accept ...

^^...

thanks for your help.....

[osan]

welcome Rich!!!
actually i wasn't sure about your code. when i was taking input from file,it was showing some wrong. but those wrong output showing right when i was giving it manually.

that why i was confused.
but at last u got AC. that's a good news.

GOOD LUCK

[Sneeze]

Yes, I agree with seadoo.
one digit of m and two digits of n are enough.

M H Rasel
CUET Old Sailor

I know the trick of finding the repeated sequence of [the last digit of m] poweres n to get the same correct answer, but could anyone please explain why the last two digits of n is enough?

[Master]

If a number N. If we can express N as

N = a*100 + b;

now if N is divisable by 4 then b must divisable by 4
and if not then b is not divisable by 4.

M H Rasel.
acmbeginner.tk