POJ 1579 Function Run Fun 记忆化递归

本题使用记忆话递归即可，也可以用动态规划DP

不过得使用三维数组，空间复杂度比较高

Function Run Fun

Time Limit:1000MS

Memory Limit:10000

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

/* Author yan
 *POJ 1579
 *Function Run Fun
*/
#include<stdio.h>
int value[61][61][61];
int w(int a,int b,int c)
{
	if(value[a+10][b+10][c+10]!=-1) return value[a+10][b+10][c+10];
	else
	{
		if(a<=0||b<=0||c<=0) return value[a+10][b+10][c+10]=1;
		else if(a>20||b>20||c>20) return value[a+10][b+10][c+10]=w(20,20,20);
		else if(a<b&&b<c) return value[a+10][b+10][c+10]=w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c);
		else return value[a+10][b+10][c+10]=w(a-1,b,c)+w(a-1,b-1,c)+w(a-1,b,c-1)-w(a-1,b-1,c-1);
	}
}
int main()
{
	int a,b,c;
	//freopen("input","r",stdin);
	while(scanf("%d %d %d",&a,&b,&c)&&(a!=-1||b!=-1||c!=-1))
	{
		memset(value,-1,sizeof(value));
		//printf("%d %d %d/n",a,b,c);
		printf("w(%d, %d, %d) = ",a,b,c);
		if(a<=0||b<=0||c<=0) printf("%d/n",1);
		else
		{
			w(a,b,c);
			printf("%d/n",value[a+10][b+10][c+10]);
		}
	}
	return 0;
}