Hdu 1331 for Memory search and hdu1331 for Memory search

Hdu 1331 for Memory search and hdu1331 for Memory search
Function Run FunTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission (s): 2586 Accepted Submission (s): 1255

Problem DescriptionWe all love recursion! Don't we?

Consider a three-parameter recursive function w (a, B, c ):

If a <= 0 or B <= 0 or c <= 0, then w (a, B, c) returns:
1

If a> 20 or B> 20 or c> 20, then w (a, B, c) returns:
W (20, 20, 20)

If a <B and B <c, then w (a, B, c) returns:
W (a, B, C-1) + w (a, B-1, C-1)-w (a, B-1, c)

Otherwise it returns:
W (A-1, B, c) + w (A-1, B-1, c) + w (A-1, B, C-1)-w (A-1, B-1, C-1)

This is an easy function to implement. the problem is, if implemented directly, for moderate values of a, B and c (for example, a = 15, B = 15, c = 15 ), the program takes hours to run because of the massive recursion.
 
InputThe input for your program will be a series of integer triples, one per line, until the end-of-file flag of-1-1-1. using the above technique, you are to calculate w (a, B, c) efficiently and print the result.
 
OutputPrint the value for w (a, B, c) for each triple.
Sample Input

1 1 12 2 210 4 650 50 50-1 7 18-1 -1 -1 

 
Sample Output

w(1, 1, 1) = 2w(2, 2, 2) = 4w(10, 4, 6) = 523w(50, 50, 50) = 1048576w(-1, 7, 18) = 1

 
SourcePacific Northwest 1999
Obviously, this question is simulated. Too many recursion times is a waste of time and cannot be passed. You can store the calculated value in the array by changing the space time, and then compute all possible calculated values. Output directly.

// Knowledge point: memory-based search uses array storage to reduce the number and time of recursive function calls. // It was made on the day of the examination. I did it again today and it turned out a little unfamiliar, B # include <stdio. h> int s [22] [22] [22]; void f () {int I, j, k; for (I = 1; I <22; ++ I) {for (j = 1; j <22; ++ j) {for (k = 1; k <22; ++ k) {if (I <j & j <k) {if (k-1 = 0) s [I] [j] [k-1] = s [I] [J-1] [k-1] = 1; if (J-1 = 0) s [I] [J-1] [k-1] = s [I] [J-1] [k] = 1; s [I] [j] [k] = s [I] [j] [k-1] + s [I] [J-1] [k-1]-s [I] [J-1] [k]; continue;} if (I = 1) s [I-1] [j] [k] = s [I-1] [J-1] [k] = s [I-1] [j] [k-1] = s [I-1] [J-1] k-1 = 1; if (j = 1) s [I-1] [J-1] [k] = s [I-1] [J-1] = 1; if (k = 1) s [I-1] [j] [k-1] = s [I-1] [J-1] [k-1] = 1; s [I] [j] [k] = s [I-1] [j] [k] + s [I-1] [J-1] [k] + s [I-1] [j] [k-1]-s [I-1] [J-1] [k-1] ;}}} int main () {int a, B, c; f (); while (~ Scanf ("% d", & a, & B, & c ),! (A =-1 & B =-1 & c =-1 )) {if (a <= 0 | B <= 0 | c <= 0) {printf ("w (% d, % d, % d) = 1 \ n ", a, B, c); continue;} if (a> 20 | B> 20 | c> 20) {printf ("w (% d, % d, % d) = % d \ n", a, B, c, s [20] [20] [20]); continue;} printf ("w (% d, % d, % d) = % d \ n", a, B, c, s [a] [B] [c]);} return 0 ;}