POJ-1579 Function Run Fun

Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13517		Accepted: 7040

Description

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. 

Input

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.

Output

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

Sample Input

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

Sample Output

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

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 int dp[21][21][21];
 8     
 9 int calw(int a, int b, int c)
10 {
11     if(a <= 0 || b <= 0 || c <= 0)
12     {
13         //[a][b][c] = 1;   //注意这里不能直接写，以为a、b、c可能超出数组范围
14         return 1;
15     }
16     if(a > 20 || b > 20 || c > 20)
17     {
18         dp[20][20][20] = calw(20, 20, 20);
19         return dp[20][20][20];
20     }
21     if(dp[a][b][c])
22         return dp[a][b][c];
23     if(a < b && b < c)
24     {
25         dp[a][b][c] = calw(a, b, c - 1) + calw(a, b - 1, c - 1) - calw(a, b-1, c);
26         return dp[a][b][c];
27     }
28     dp[a][b][c] = calw(a-1, b, c) + calw(a - 1, b - 1, c) + calw(a - 1, b, c - 1) - calw(a - 1, b - 1, c - 1);
29     return dp[a][b][c];
30 }
31 
32 int main()
33 {
34     int a, b, c;
35     while(cin >> a >> b >> c && a != -1 || b != -1 || c != -1)
36     {
37         printf("w(%d, %d, %d) = %d\n",a, b, c, calw(a, b, c));
38     }
39     return 0;
40 }