Fibonacci Check-up

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 42 Accepted Submission(s): 27

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 Fibonacci Check-up-LMLPHP , is the combination number. The answer mod m (the total number of alpc team members) is just your alpc-number.

Input

First line is the testcase T.
Following T lines, each line is two integers n, m ( 0<= n <= 10^9, 1 <= m <= 30000 )

Output

Output the alpc-number.

Sample Input

Sample Output

1

3

Source

2009 Multi-University Training Contest 5 - Host by NUDT

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gaojie

/*

题意：给出你公式，让你求( 求和C(k,n)F(k) )%m

初步思路：没思路，先打表

            得到：

            0

            1

            3

            8

            21

            55

            144

            377

            987

            2584

            6765

            17711

            46368

            121393

            317811

            832040

            2178309

            5702887

            14930352

            39088169

            102334155

            能得出来，G(0)=0;

                      G(1)=1;

                      G(n)=3*G(n-1)-G(n-2);

            然后用矩阵快速幂求出结果

            最重要的构造矩阵还没学线性代数的我只能试着推出来：

            |G(n) G(n-1)| | 3     1   |

            | 0    0    |*| -1    0   |

#总结：板没调好，WA了两发

*/

#include<bits/stdc++.h>

using namespace std;

int n,mod;

int t;

/********************************矩阵模板**********************************/

class Matrix {

    public:

        int a[][];

        int n;

        void init(int x) {

            memset(a,,sizeof(a));

            if(x){

                a[][]=;

                a[][]=;

                a[][]=-;

                a[][]=;

            }else{

                a[][]=;

                a[][]=;

                a[][]=;

                a[][]=;

            }

            n=;

        }

        Matrix operator +(Matrix b) {

            Matrix c;

            c.n = n;

            for (int i = ; i < n; i++)

                for (int j = ; j < n; j++)

                    c.a[i][j] = (a[i][j] + b.a[i][j]) % mod;

            return c;

        }

        Matrix operator +(int x) {

            Matrix c = *this;

            for (int i = ; i < n; i++)

                c.a[i][i] += x;

            return c;

        }

        Matrix operator *(Matrix b)

        {

            Matrix p;

            p.n = b.n;

            memset(p.a,,sizeof p.a);

            for (int i = ; i < n; i++)

                for (int j = ; j < n; j++)

                for (int k = ; k < n; k++)

                    p.a[i][j] = (p.a[i][j] + (a[i][k]*b.a[k][j])%mod) % mod;

            return p;

        }

        Matrix power_1(int t) {

            Matrix ans,p = *this;

            ans = p;

            while (t) {

                if (t & )

                    ans=ans*p;

                p = p*p;

                t >>= ;

            }

            return ans;

        }

        Matrix power_2(Matrix a,Matrix b,int x){

            while(x){

                if(x&){

                    b=a*b;

                }

                a=a*a;

                x>>=;

            }

            return b;

        }

};

/********************************矩阵模板**********************************/

Matrix unit,init;

int main(){

    // freopen("in.txt","r",stdin);

    scanf("%d",&t);

    while(t--){

        scanf("%d%d",&n,&mod);

        unit.init();//存放G(n)的矩阵

        init.init();//子矩阵

        if(n==){

            printf("0\n");

            continue;

        }else if(n==){

            printf("%d\n",%mod);

            continue;

        }

        init=init.power_1(n-);

        unit=unit*init;

        // cout<<mod<<endl;

        // for(int i=0;i<2;i++){

            // for(int j=0;j<2;j++){

                // cout<<init.a[i][j]<<" ";

            // }

            // cout<<endl;

        // }

        printf("%d\n",(unit.a[][]+mod)%mod);

    }

    return ;

}

/*

附上打表小程序

*/

#include<bits/stdc++.h>

using namespace std;

long long Sums(long long n,long long k) //函数定义

{

    long long sum = ,N=,K=,M=;

    if(k >  && k<=n)

    {

        for(long long i = ;i<=n;i++)

        {

            N = N * i;

        }

        for(long long j = ;j <= k;j++)

        {

            K = K * j;

        }

        for(long long h = ;h <= n-k;h++)

        {

            M = M * h;

        }

        sum=N/(K*M);

        return sum;

    }

    else

        return ;

}

int main(){

    //freopen("in.txt","r",stdin);

    long long f[];

    f[]=;

    f[]=;

    for(long long i=;i<=;i++)

        f[i]=f[i-]+f[i-];

    for(long long n=;n<=;n++){

        long long cur=;

        for(long long k=;k<=n;k++){

            long long cnk=Sums(n,k);

            //cout<<"C("<<k<<","<<n<<")="<<cnk<<endl;

            cur+=cnk*f[k];

        }

        cout<<cur<<endl;

    }

    return ;

}