POJ 1579 Function Run Fun【递归转循环】

最新推荐文章于 2018-08-26 19:15:58 发布

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本文介绍了一个三参数递归函数w(a,b,c)的高效实现方法。通过预先计算并存储递归结果避免重复计算，显著提高了运行效率。文章提供了一个AC代码示例，展示了如何处理边界条件，并给出了一些样本输入和输出。

Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18342		Accepted: 9376

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

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

AC代码：

#include<cstdio>

int w[22][22][22];

int main()
{
	int a,b,c;
	for(int i=0;i<21;++i) {
		for(int j=0;j<21;++j) {//w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1) 
			for(int k=0;k<21;++k) {
				if(!i||!j||!k) 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];
				//printf("%d %d %d %d\n",i,j,k,w[i][j][k]);
			}
		}
	}
	while(scanf("%d%d%d",&a,&b,&c),~a||~b||~c)
	{
		int ans=0;
		if(a<=0||b<=0||c<=0) ans=1;
		else if(a>20||b>20||c>20) ans=w[20][20][20];
		else ans=w[a][b][c];
		printf("w(%d, %d, %d) = %d\n",a,b,c,ans); 
	}
	return 0;
}