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

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int dp[21][21][21];

int calw(int a, int b, int c)
{
if(a <= 0 || b <= 0 || c <= 0)
{
//[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
return 1;
}
if(a > 20 || b > 20 || c > 20)
{
dp[20][20][20] = calw(20, 20, 20);
return dp[20][20][20];
}
if(dp[a][b][c])
return dp[a][b][c];
if(a < b && b < c)
{
dp[a][b][c] = calw(a, b, c - 1) + calw(a, b - 1, c - 1) - calw(a, b-1, c);
return dp[a][b][c];
}
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);
return dp[a][b][c];
}

int main()
{
int a, b, c;
while(cin >> a >> b >> c && a != -1 || b != -1 || c != -1)
{
printf("w(%d, %d, %d) = %d\n",a, b, c, calw(a, b, c));
}
return 0;
}