poj-1579 fuction

Function Run Fun

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 21720 Accepted: 10807

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
Source

Pacific Northwest 1999

#include<cstring>
#include<iostream>
#include<stdio.h>
using namespace std;
int dp[100][100][100];
int dfs(int a,int b,int c)
{
	if(a<=0||b<=0||c<=0)
	return 1;
	if(dp[a][b][c]) return dp[a][b][c];
	if(a>20||b>20||c>20) return 1048576;
	if(a < b && b < c) return dp[a][b][c] = dfs(a,b,c-1) + dfs(a,b-1,c-1) - dfs(a,b-1,c);
	else return dp[a][b][c] = dfs(a-1,b,c) + dfs(a-1,b-1,c) + dfs(a-1,b,c-1) - dfs(a-1,b-1,c-1);
}
int main()
{
	int a,b,c;
	while(cin>>a>>b>>c)
	{
		if(a==-1&&b==-1&&c==-1)
		break;
		else
		printf("w(%d, %d, %d) = %d\n",a,b,c,dfs(a,b,c));
	}

	return 0;
}

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有几点注意：
1、对于题目中所给出的条件，利用简单递归即可。
2、在函数中，把每次的a,b,c所代表的值用dp[a][b][c]储存起来，以便下次调用的时候节省时间（这就是最简单的记忆化搜索），不然只能tle了。
3、输出的格式一定要写对，最好从题中复制粘贴！（不然迟早要吃亏！！！）

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