Fibonacci Check-up
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1436 Accepted Submission(s): 825
Problem Description
Every ALPC has his own alpc-number just like alpc12, alpc55, alpc62 etc.
As more and more fresh man join us. How to number them? And how to avoid their alpc-number conflicted?
Of course, we can number them one by one, but that’s too bored! So ALPCs use another method called Fibonacci Check-up in spite of collision.
First you should multiply all digit of your studying number to get a number n (maybe huge).
Then use Fibonacci Check-up!
Fibonacci sequence is well-known to everyone. People define Fibonacci sequence as follows: F(0) = 0, F(1) = 1. F(n) = F(n-1) + F(n-2), n>=2. It’s easy for us to calculate F(n) mod m.
But in this method we make the problem has more challenge. We calculate the formula
, is the combination number. The answer mod m (the total number of
alpc team members) is just your alpc-number.
As more and more fresh man join us. How to number them? And how to avoid their alpc-number conflicted?
Of course, we can number them one by one, but that’s too bored! So ALPCs use another method called Fibonacci Check-up in spite of collision.
First you should multiply all digit of your studying number to get a number n (maybe huge).
Then use Fibonacci Check-up!
Fibonacci sequence is well-known to everyone. People define Fibonacci sequence as follows: F(0) = 0, F(1) = 1. F(n) = F(n-1) + F(n-2), n>=2. It’s easy for us to calculate F(n) mod m.
But in this method we make the problem has more challenge. We calculate the formula

Input
First line is the testcase T.
Following T lines, each line is two integers n, m ( 0<= n <= 10^9, 1 <= m <= 30000 )
Following T lines, each line is two integers n, m ( 0<= n <= 10^9, 1 <= m <= 30000 )
Output
Output the alpc-number.
Sample Input
2 1 30000 2 30000
Sample Output
1 3
#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
const int N=2;
struct Matrix{
LL m[N][N];
};
Matrix A={
1,1,
1,0
};
Matrix I={
1,0,
0,1
};
Matrix multi(Matrix a,Matrix b,int mod)
{
Matrix c;
for(int i=0;i<N;i++)
{
for (int j=0;j<N;j++)
{
c.m[i][j]=0;
for(int k=0;k<N;k++)
{
c.m[i][j]+=a.m[i][k]*b.m[k][j]%mod;
}
c.m[i][j]%=mod;
}
}
return c;
}
Matrix power(Matrix A,int k,int mod)
{
Matrix ans=I,p=A;
while(k)
{
if(k&1)
{
ans=multi(ans,p,mod);
k--;
}
k>>=1;
p=multi(p,p,mod);
}
return ans;
}
int main()
{
int n,mod;
LL T;
cin>>T;
while(T--)
{
scanf("%d%d",&n,&mod);
if(n==0)
printf("0\n");
else{
n*=2;
Matrix ans=power(A,n-1,mod);
printf("%I64d\n",ans.m[0][0]);
}
}
return 0;
}