解题报告1579

Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6707		Accepted: 3651

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

题目描述了一个递归过程，要求编程实现求此递归函数的功能。

如果直接按照递归定义来写，肯定会TLE，因为递归的过程中会产生很多的递归分支，而递归函数的调用又会完成很多不必要的重复工作。

这里的思想是预存储，典型的用空间换时间。

如果直接按照递归定义计算，计算时严格按照递归定义回溯，每次计算可能都会经历很多次递归调用，可能会重复计算。如果将所有结果都预处理放到数组，递归的过程中，大变量结果由小变量结果得出，只需遍历一遍，便可算出所有结果，而且每个结果的计算只需按照递归定义计算一层。

注意到，递归的过程中始终是向着变量减小的方向，于是可以由变量由小到大遍历一次，算出变量所有情况下的函数结果，存储在数组中。真正求解时，直接返回结果。

代码如下：

#include <iostream> using namespace std; int main() { int result[21][21][21]; result[0][0][0] = 1; for (int i = 0;i < 21;i++) { for (int j = 0;j < 21;j++) { for (int k = 0;k < 21;k++) { if(!i||!j||!k) result[i][j][k] = result[0][0][0]; else if(i<j&&j<k) result[i][j][k] = result[i][j][k-1] + result[i][j-1][k-1] - result[i][j-1][k]; else result[i][j][k] = result[i-1][j][k] + result[i-1][j-1][k] + result[i-1][j][k-1] - result[i-1][j-1][k-1]; } } } int a,b,c; while (1) { cin>>a>>b>>c; if(!(a+1)&&!(b+1)&&!(c+1)) break; cout<<"w("<<a<<", "<<b<<", "<<c<<") = "; if(a>20&&b>20&&c>20) cout<<result[20][20][20]; else if(a<=0||b<=0||c<=0) cout<<1; else cout<<result[a][b][c]; cout<<endl; } }