Cod sursa(job #732632)

Utilizator a_h1926Heidelbacher Andrei a_h1926 Data 10 aprilie 2012 18:58:12
Problema Pod Scor 85
Compilator cpp Status done
Runda Arhiva de probleme Marime 4.12 kb
/*#include <cstdio>
#include <algorithm>
#include <cstring>

#define KMax 22
#define MMax 1005
#define Mod 9901

using namespace std;

int N, K, M[MMax], DP[KMax];

class Matrix
{
    public: int x[KMax][KMax];

    Matrix ()
    {
        memset (x, 0, sizeof (x));
    }

    Matrix (int v)
    {
        memset (x, 0, sizeof (x));
        for (int k=1; k<=K; ++k) x[k][k]=v;
    }

    inline void operator = (Matrix y)
    {
        memcpy (x, y.x, sizeof (y.x));
    }

    inline Matrix operator * (Matrix y)
    {
        Matrix z (0);
        for (int i=1; i<=K; ++i)
        {
            for (int j=1; j<=K; ++j)
            {
                for (int k=1; k<=K; ++k)
                {
                    z.x[i][j]+=(x[i][k]*y.x[k][j]);
                    if (z.x[i][j]>=Mod) z.x[i][j]%=Mod;
                }
            }
        }
        return z;
    }
};

Matrix Z (0);

void BuildZ ()
{
    Z.x[1][K]=Z.x[K][K]=1;
    for (int i=2; i<=K; ++i)
    {
        Z.x[i][i-1]=1;
    }
}

void InitDP ()
{
    for (int i=1; i<=K; ++i) DP[i]=1;
}

inline Matrix Pow (Matrix &X, int P)
{
    Matrix R (1);
    while (P>0)
    {
        if (P%2==0) X=(X*X), P/=2;
        else R=(R*X), --P;
    }
    return R;
}

inline void Compute (int P)
{
    Matrix X; X=Z;
    X=Pow (X, P);
    int NewDP[KMax]; memset (NewDP, 0, sizeof (NewDP));
    for (int i=1; i<=K; ++i)
    {
        for (int j=1; j<=K; ++j)
        {
            NewDP[i]=(NewDP[i]+DP[j]*X.x[j][i])%Mod;
        }
    }
    for (int i=1; i<=K; ++i) DP[i]=NewDP[i];
}

void Solve ()
{
    BuildZ ();
    sort (M+1, M+M[0]+1);
    int P=K;
    for (int i=1; i<=M[0]; ++i)
    {
        Compute (M[i]-P);
        DP[K]=0; P=M[i];
    }
}

void Read ()
{
    freopen ("pod.in", "r", stdin);
    int Broken; scanf ("%d %d %d", &N, &Broken, &K);
    InitDP ();
    for (int i=1; i<=Broken; ++i)
    {
        int P; scanf ("%d", &P); ++P;
        if (P<=K)
        {
            for (int j=P; j<=K; ++j) DP[j]=0;
        }
        else M[++M[0]]=P;
    }
    M[++M[0]]=N+2;
}

void Print ()
{
    freopen ("pod.out", "w", stdout);
    if (N<K) printf ("%d\n", DP[N]);
    else printf ("%d\n", DP[K-1]);
}

int main()
{
    Read ();
    Solve ();
    Print ();
    return 0;
}
*/

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define MOD 9901

int n,m,k,p;
int v[1006];
int rez[56];
int rez2[56];

struct matrix
{
    int mt[34][34];

    matrix(){
    memset(mt,0,sizeof(mt));}

    matrix(int c)
    {
        int i;
        memset(mt,0,sizeof(mt));
        for(i=1;i<=k;i++)
            mt[i][i]=c;
    }

    matrix operator*(matrix b)
    {
        int i,j,t;
        matrix c(0);
        for(t=1;t<=k;t++)
            for(i=1;i<=k;i++)
                for(j=1;j<=k;j++)
                {
                    c.mt[i][j]+=mt[i][t]*b.mt[t][j];
                    if(c.mt[i][j]>=MOD)
                        c.mt[i][j]%=MOD;
                }
        return c;
    }
};

matrix mat,mat2;

matrix rid_log(matrix a,int put)
{
    if(!put)
    {
        matrix z(1);
        return z;
    }
    if(put==1)
        return a;
    matrix b=rid_log(a,put/2);
    b=b*b;
    if(put&1)
        b=b*a;
    return b;
}


int main ()
{
    int i,j,t;

    freopen("pod.in","r",stdin);
    freopen("pod.out","w",stdout);

    scanf("%d%d%d",&n,&m,&k);

    for(i=1;i<=m;i++)
        scanf("%d",&v[i]);
    v[++m]=n;

    for(i=1;i<k;i++)
        mat.mt[i][i+1]=1;
    mat.mt[k][1]=mat.mt[k][k]=1;

    rez[k]=1;
    sort(v+1,v+m+1);
    for(i=1;i<=m;i++)
    {
        mat2=rid_log(mat,v[i]-p);
        memset(rez2,0,sizeof(rez2));
        for(j=1;j<=k;j++)
            for(t=1;t<=k;t++)
            {
                rez2[j]+=mat2.mt[j][t]*rez[t];
                if(rez2[j]>=MOD)
                    rez2[j]%=MOD;
            }
        if(i<m)
            rez2[k]=0;

        for(j=1;j<=k;j++)
            rez[j]=rez2[j];
        p=v[i];
    }
    printf("%d\n",rez[k]);
    return 0;
}