OJ Board

already correct that! Thanks

but my program is still taking to much time....

Is there any method to make it more efficient?

thanks before

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

Here is a part from your code

;
for(i=0;i<n;i++)
{ temp=(int)m%10;
digit*=(int)temp;
digit=digit%10;
}

I guess you should realize that this loop will take far too long for the worst case of n=10^101-1.

Allright here's a hint......
Do you really have to carry out the multiplications to find the last digit ?

Example : 9^1 = 9
9^2 = ..1
9^3 = ..9
9^4 = ..1

etc. where .. denotes digits other than last digit.

There's a pattern.....

thanks for all your explenation, I really appriciate it.

I've tried almost all cases but the judge gave me WA.....

[c]
#include<stdio.h>
#include<string.h>

void main()
{ char m[101],n[101];
int arr[4];
int p1,p2,m1,n1,i,t;
while(1)
{ m1=0;
n1=0;
scanf("%s %s",m,n);
if(m[0]=='0' && n[0]=='0') break;
p1=strlen(m);
p2=strlen(n);
m1=m[p1-1]-48;

n1+=n[p2-1]-48;
if(p2>1)
n1+=(n[p2-2]-48)*10;

m1=m1%10;
n1=(n1-1)%4;
t=1;
for(i=0;i<4;i++)
{ t*=m1;
arr=t;
}
if(p2==1 && n[p2-1]=='0')
arr[0]=1;
printf("%d\n",arr[n1]%10);
}
}
[/c]

your program looks OK .......

I don't know what's wrong about my code.

I have test a lot, thus don't got lost.

Who can help me, thanks in advance.

[cpp]#include <iostream>
#include <string.h>

using namespace std;

int const Max1=101;
int const Max2=4; // at most 4 different case

char A[Max1];
char B[Max1];
int S[Max2];

void main()
{
int i;
for(i=0;i<Max1;i++) // clear stage
A=B=NULL;
for(i=0;i<Max2;i++)
S=0;
while(1)
{
cin>>A>>B;
int a=0;
while(A[a]!=NULL)
a++;
int length_m=a;
a--;
int m=A[a]-'0';
a=0;
while(B[a]!=NULL)
a++;
int length_n=a;
a-=2;
int n;
if(B[a]>'0'&&B[a]<='9')
n=(B[a]-'0')*10+(B[a+1]-'0');
else
n=(B[a+1]-'0');
if(m==0&&n==0&&length_m==1&&length_n==1) break;
else if(m==0&&n==0&&length_m>1)
{
cout<<'0'<<endl;
continue;
}
else if(n==0)
{
cout<<'1'<<endl;
continue;
}
int cnt=m;
int k=0;
int i,j;
for(i=0;i<Max2;i++)
{
if(m>0) m%=10;
if(!S[0]&&m!=S[0]) S[0]=m;
else if(!S[1]&&m!=S[1]) S[1]=m;
else if(!S[2]&&m!=S[2]) S[2]=m;
else if(!S[3]&&m!=S[3]) S[3]=m;
m*=cnt;
}
for(j=0;j<Max2;j++)
if(S[j]) continue;
cout<<S[(n-1)%j]<<endl;
for(a=0;a<Max1;a++) // clear stage
A[a]=B[a]=NULL;
for(a=0;a<Max2;a++)
S[a]=0;
}
}[/cpp]

Anyone help me!

I'm crazy!!!!

Zhao Le wrote:Anyone help me!

I'm crazy!!!!

Try this input:
10 0
0 0

Try this input:
10 0
0 0

I have solved this bug.

But when I sumbit to the judge, I was suprised to see it is still WA.

So there is something wrong with my code.

Here is my new code.

[cpp]#include <iostream.h>

int const Max1=101;
int const Max2=4; // at most 4 different case

char A[Max1];
char B[Max1];
int S[Max2];

void main()
{
int i;
for(i=0;i<Max1;i++) // clear status
A=B=NULL;
for(i=0;i<Max2;i++)
S=0;
while(1)
{
cin>>A>>B;
int a=0;
while(A[a]!=NULL)
a++;
int length_m=a;
a--;
int m=A[a]-'0'; // last digit of M
a=0;
while(B[a]!=NULL)
a++;
int length_n=a;
a-=2;
int n;
if(B[a]>'0'&&B[a]<='9')
n=(B[a]-'0')*10+(B[a+1]-'0'); // last two digit of N
else
n=(B[a+1]-'0');// last digit of N
if(m==0&&n==0&&length_m==1&&length_n==1) break; // case 0 0
else if(n==0&&length_n==1) // case X 0
{
cout<<'1'<<endl;
continue;
}
int cnt=m;
int k=0;
int i,j;
for(i=0;i<Max2;i++) // At most four cases
{
if(m>0) m%=10;
if(!S[0]&&m!=S[0]) S[0]=m;
else if(!S[1]&&m!=S[1]) S[1]=m;
else if(!S[2]&&m!=S[2]) S[2]=m;
else if(!S[3]&&m!=S[3]) S[3]=m;
m*=cnt;
}
for(j=0;j<Max2;j++)
if(S[j]) continue;
cout<<S[(n-1)%j]<<endl;
for(a=0;a<Max1;a++) // clear status
A[a]=B[a]=NULL;
for(a=0;a<Max2;a++)
S[a]=0;
}
}[/cpp]

Can someone give me test case!

Thanks in advance.

Zhao Le wrote:Can someone give me test case!

Thanks in advance.

Here's another one:

12 100
0 0

Finally, I got AC.

There is not words can express my thanks.

wud anyone plz help me to understand the logic of powers et all?
i took the last digit of base and the last two digits of power.it works fine for many inputs,but in case of 2^100 it gives the output 0,whereas the correct output is 6.plz help me

Well, what's the last digit of 2^99? .. If you multiply that digit by two (thus getting 2^100), you will get 6, and definitely not 0 (unless 2^99 had a last digit of 0, and 2^98, etc, which you know is not the case).

Now, you should ask yourself, is taking the last two digits enough when the last two digits are equivalent to '00'?

Eric, can you post a huge test input/output file? I'm getting WA all the times, but my solution DOES passes all the inputs here.