2013
11-09

# Function Run Fun

We 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.

The 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.

Print the value for w(a,b,c) for each triple.

1 1 1
2 2 2
10 4 6
50 50 50
-1 7 18
-1 -1 -1

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


import java.util.Scanner;
import java.util.StringTokenizer;

public class Main {

public static void main(String[] args) throws Exception{

int [][][]w = new int[21][21][21];
for(int i=0;i< 21;i++){
for(int j=0;j< 21;j++){
for(int k=0;k< 21;k++){
if(i==0||j==0||k==0)
w[i][j][k] = 1;
else{
if(i< j&&j< k)
w[i][j][k] = w[i][j][k-1] + w[i][j-1][k-1] - w[i][j-1][k];
else
w[i][j][k] = w[i-1][j][k] + w[i-1][j-1][k] + w[i-1][j][k-1] - w[i-1][j-1][k-1];
}
}
}
}

Scanner  in = new Scanner(System.in);
String input = new String();
while(!(input=in.nextLine()).equals("-1 -1 -1")) {
StringTokenizer toke = new StringTokenizer(input);
int a = Integer.parseInt(toke.nextToken());
int b = Integer.parseInt(toke.nextToken());
int c = Integer.parseInt(toke.nextToken());

int result = 0;
if (a<=0||b<=0||c<=0)
result = 1;
else {
if(a>20||b>20||c>20)
result = w[20][20][20];
else
result = w[a][b][c];
}
System.out.println("w("+a+", "+b+", "+c+") = "+result);
}
}

}

1. 为什么for循环找到的i一定是素数叻，而且约数定理说的是n=p1^a1*p2^a2*p3^a3*…*pk^ak，而你每次取余都用的是原来的m，也就是n