C-Language algorithm: Perfecting the K-order Fibonacci sequence of the year

The following is the log replay for the sophomore time:
"The topic extends to the K -order,

kThe order Fibonacci sequence, 1 order (i.e.k=1):1,1,1,1,1,1,1、......

a0=a[1-1]=1,a1=1,a2=1,a3=1,a4=1,a5=1,a6=1 ...

3Order (k=3):0,0,1,1,2,4,7、、、、、

A0=0,a1=0,a2=a[3-1]=1,a3=0+0+1=1,a4=0+1+1=2,a5=1+2+4=7

4Order:0,0,0,1,1,2,4,8, the, -...

a0=0,a1=0,a2=0,a3=a[4-1]=1,a4=1,a5=2,a6=4 ... a[8]=1+2+4+8=15 ...

The problem is generalized and it can be seen that:

k-2 0 ;

Section k-1 the value of the item is 1 ;

Sectionkto the2k-1the value of the item is2of theNThe Second party (also the formerkitems and But at the same time with2of theNThe same as the second-party value);

Section 2k The value after the item is the previous k items and;

Web-based method of ' recursion ', and only correspondence 2 step, not extended to K step, the younger brother also skilled ' recursion ', or continue to use the order method.

c

Status fibonacci (int k, int m, int &f)/*  the value of item m of the K-order Fibonacci sequence f     */{ int a[100],i,y,z=0;     /* defines an array with an element number of 100. */ if (k<=1| | M&LT;0) return error;    /* returns an error warning */ else if (m<k-1) If the input parameter is not canonical f=0;       /* constant equals data segment processing */ else if (m==k-1) f=1; else if (m<=2*k-1) F=pow ( 2,M-K);     /* two-segment processing */ else                   /*k stage processing, time complexity */ { for (y=0;y<k;y++)        /* assigns a value */ a[y]=pow (2,y) to the K array element with a for loop;  for (i=2*k;i<=m;i++)  { &NBSP;A[K]=0;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;/*A[K] */ for (y=0;y<k;y++)       /* embedded in the For loop body inside the For loop body */ { a[k] Each time the loop is cleared 01 times =a[k]+a[y];        /* K-Item plus */ }   a[z]=a[k];   /* Each move to the value of the smallest element of the eradication, change to the maximum value, do not have to assign a value, save a large amount of time, again added "output A[K] value" can achieve the first m-k and output */  z++;       /*z each time you add 1, make sure to overwrite the smallest item with the largest item of the top K item */  if  (z==k) z=0;  /* when z==k, Z zeroing */  }   f=a[k];      /* output: The value of item m of the K-order Fibonacci sequence number F */ } return  ok;}

The following is an optimized algorithm that has a faster processing time in the Obline judge.

Status fibonacci (int k, int m, int &f)/*  the value of item m of the K-order Fibonacci sequence f     */{ long a[300], i, y, z=0;                     /* defines an array with an element number of 300, The limit in long is less than the No. 300 number. */ if (k <= 1| | M&NBSP;&LT;&NBSP;0) return error;    /* returns an error warning */&NBSP;ELSE&NBSP;IF if the input parameter is not canonical (m  &LT;&NBSP;K-1) f=0;                        /* constant equals data segment processing */ else if (m == k-1) f=1;  Else for (i=0; i<k; i++) a[i]=0;  a[k]=1;                              /* input k 0 and A[k]=1 initial digital */ for (y=k+1; y<=m; y++)  a[y]=a[y-1]*2-a[y-k];          /* Law: The number of y is equal to y-1, multiplied by 2 minus the y-k number */   f=a[y-1];                             /* output: The value of item m of the K-order Fibonacci sequence number f */  } return ok;}

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C-Language algorithm: Perfecting the K-order Fibonacci sequence of the year