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  • UVA 10870 Recurrences

    矩阵快速幂。

    矩阵很容易构造:

    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<vector>
    #include<algorithm>
    using namespace std;
    
    int d,n;
    long long m;
    long  long F[20],D[20];
    
    struct Matrix
    {
        long long A[20][20];
        int R, C;
        Matrix operator*(Matrix b);
    };
    
    Matrix X, Y, Z;
    
    Matrix Matrix::operator*(Matrix b)
    {
        Matrix c;
        memset(c.A, 0, sizeof(c.A));
        int i, j, k;
        for (i = 1; i <= R; i++)
            for (j = 1; j <= C; j++)
                for (k = 1; k <= C; k++)
                    c.A[i][j] = (c.A[i][j] + (A[i][k] * b.A[k][j])%m)%m;
        c.R=R; c.C=b.C;
        return c;
    }
    
    void init()
    {
        n = n - 1;
        memset(X.A,0,sizeof X.A);
        memset(Y.A,0,sizeof Y.A);
        memset(Z.A,0,sizeof Z.A);
    
        for(int j=1;j<=d;j++) Z.A[1][j]=F[j]; Z.R = 1; Z.C = d;
    
        for(int i=1;i<=d;i++) Y.A[i][i]=1; Y.R = d; Y.C = d;
    
        for(int j=1;j<=d-1;j++) X.A[j+1][j]=1;
        for(int i=1;i<=d;i++) X.A[i][d]=D[d-i+1];
         X.R = d; X.C = d;
    }
    
    void work()
    {
        while (n)
        {
            if (n % 2 == 1) Y = Y*X;
            n = n >> 1;
            X = X*X;
        }
        Z = Z*Y;
    
        printf("%lld
    ", Z.A[1][1]);
    }
    
    void read()
    {
        for(int i=1;i<=d;i++) scanf("%lld",&D[i]);
        for(int i=1;i<=d;i++) scanf("%lld",&F[i]);
    }
    
    int main()
    {
        while(~scanf("%d%d%lld",&d,&n,&m))
        {
            if(d==0&&n==0&&m==0) break;
            read();
            init();
            work();
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zufezzt/p/5237231.html
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