POJ 1579 Function Run Fun

Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19515		Accepted: 9866

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

题目比较简单，但是虽然题目给了你递归的表达式，但是我们不能直接使用，不然的话，肯定超时，

我的做法相对简单，直接，用for循环打表，O(n^3)。但是比较坑的地方在于，要完全按照他递归的写法来循环

#include<cstdio>
#include<algorithm>
#include<set>
#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
int f[25][25][25];
int main ()
{
int a,b,c;
memset(f,0,sizeof(0));
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)
f[i][j][k]=1;
else if(i<j&&j<k)
f[i][j][k]=f[i][j][k-1]+f[i][j-1][k-1]-f[i][j-1][k];
else
f[i][j][k]=f[i-1][j][k]+f[i-1][j-1][k]+f[i-1][j][k-1]-f[i-1][j-1][k-1];
}
while(~scanf("%d%d%d",&a,&b,&c))
{
if(a==-1&&b==-1&&c==-1) break;
int d=a,e=b,h=c;
if(a<=0||b<=0||c<=0) a=0,b=0,c=0;//这里的条件判断不能写反，题目是先判断是否小于0再判断是否大于20的
else if(a>20||b>20||c>20) a=20,b=20,c=20;

printf("w(%d, %d, %d) = %d\n",d,e,h,f[a][b][c]);
}

}