"100 个台阶"问题的 4 种解法

 

/*

问题描述:

 

100 个台阶, 每次可以走 1 或 2 或 3 个台阶, 走完这 100 个台阶共有多少种走法?

 

基本思想:

f(1) = 1;

f(2) = 2;

f(3) = 4;

f(n) = f( n-1 ) 最后一步为 1

      + f( n-2 ) 最后一步为 2

      + f( n-3 ) 最后一步为 3

*/

 

 

 

double recursion1( int n )

{

    if( n == 1 )

       return 1;

    else if( n == 2 )

       return 2;

    else if( n == 3 )

       return 4;

    else

    {

       printf("%5d", n );

       return recursion1( n-1 ) + recursion1( n-2 ) + recursion1( n-3 );

    }

}

 

 

double recursion2( int n )

{

    /*

    算法思想:

       将已经算出的 数保存在 cash[] 中, 每次先在 cash[] 中查找, 如果找到则不再递归, 直接返回结果.

    */

    static double cash[100] = {0,1,2,4};  

    if( cash[n] != 0 )

       return cash[n];

    else

    {

       printf("%5d", n );

       cash[n] = recursion2( n - 3 ) + recursion2( n-2 ) + recursion2( n-1 );

       return cash[n];

    }

}

 

double no_recursion1( int n )

{

    if( n == 1 )

       return 1;

    else if( n == 2 )

       return 2;

    else if( n == 3 )

       return 4;

   

    double t_3 = 1; // record f( n-3 )

    double t_2 = 2; // record f( n-2 )

    double t_1 = 4; // record f( n-1 )

    double sum = 0;

    for( int i = 4; i <= n; i ++ )

    {

       // f(n) = f( n-1 ) + f( n-2 ) + f( n-3 )

       sum = t_1 + t_2 + t_3;

 

       // update f( n-1 ), f( n-2 ), f( n-3 )

       t_3 = t_2;

       t_2 = t_1;

       t_1 = sum;

    }

    return sum;

}

 

double ff( int end, int beg = 1 )

{

    if( end == 0 || end == 1)

       return 1;

 

    double ss = 1;

    for( int i = beg; i <= end ; i ++ )

       ss *= i;

    return ss;

}

 

double no_recursion2( int nn )

{

    /*

    算法思想:

       找出满足 x + 2y + 3z == nn 的所有 x, y, z, 之后问题变成一个重集 B 上的全排列问题:

       其中 B = { x.1, y.2, z.3 },

       公式为:     nn!/( x! * y! * z! )

       (此处对阶乘算法没有进一步优化)

    */

    double sum = 0;

    for( int z = 0; z < 34; z ++ )

    {

       for( int y = 0; y < 51; y ++ )

       {

           for( int x = 0; x < 101; x ++ )

           {

              if( x + 2*y + 3*z != nn )

                  continue;

 

              // x + 2y + 3z == 100

              double t1 = ff( x );

              double t2 = ff( y );

              double t3 = ff( z );

              double t4 = ff( x + y + z );

              double tem = t4 / ( t1 * t2 * t3 );

              sum += tem;

              printf("%5d%5d%5d%50.0f/n", x, y, z, tem );

           }

       }

    }

    return sum;

}

 

 

int main()

{

   

    //double sum = recursion1( 25 ); // 25 is the MAX number that I can bear!

    double sum = recursion2( 100 );

    //double sum = no_recursion1( 100 );

    //double sum = no_recursion2( 100 );

   

    printf("/n/nsum is: %.0f", sum );

    printf("/n/nsum is: %.20G/n", sum );

 

}