模板-->常系数线性齐次递推(矩阵快速幂)

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相关链接

• 所有模板的快速链接
• Matrix模板
• poj_2118_Firepersons,my_ac_code

简单的测试

None

代码模板

/*
 * TIME COMPLEXITY:O(n^3log(t))
 * PARAMS:
 *      a       The constant array.
 *      b       The initial array.
 *      n       The length of array.
 *      t       The iterator's value.
 *
 *  MATRIX M:
 *  0       1       0       ...     0
 *  0       0       1       ...     0
 *  ...             ...         ...
 *  0       0       0       ...     1
 * a[n-1]  a[n-2]  a[n-3]   ...    a[0]    But,here we use M.a[last row][i]=a[i],so inverse a before you call this function.
 *
 * MATRIX F
 * b[0]
 * b[1]
 *  .
 *  .
 *  .
 * b[n]   As we call M*F,the first row will gone and insert a new b[n+1]=M.a[last_row]*F.a[first_column] at the last row position.Until t is reached.
 */
int solve(int a[],int b[],int n,int t){
    Matrix M,F,E;
    M.clear(),F.clear(),E.clear();
    M.n=M.m=n;
    E.n=E.m=n;
    F.n=n,F.m=1;
    for(int i=0;i<n-1;i++)
        M.a[i][i+1]=1;
    for(int i=0;i<n;i++){
        //notice:Maybe,first you should inverse a array before you call this function.
        M.a[n-1][i]=a[i];
        F.a[i][0]=b[i];
        E.a[i][i]=1;
    }
    if(t<n)
        return F.a[t][0];
    //Here is fast Exponentiation.
    for(t-=n-1;t;t/=2){
        if(t&1)
            E=M*E;
        M=M*M;
    }
    F=E*F;
    return F.a[n-1][0];
}