Function Run Fun(递归转递推)

Link：http://acm.hdu.edu.cn/showproblem.php?pid=1579

Problem：

Function Run Fun

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2223    Accepted Submission(s): 1170

Problem 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

冬练三九之一

 

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#include<iostream>
#include<algorithm>
using  namespace std;
int num[22][22][22];
int main()
{
int i,j,k,a,b,c;
for(i=0;i<21;i++)
{
for(j=0;j<21;j++)
{
for(k=0;k<21;k++)
{
if(!i||!j||!k)
num[i][j][k]=1;
else if(i<j&&j<k)
num[i][j][k]=num[i][j][k-1]+num[i][j-1][k-1]-num[i][j-1][k];
else
num[i][j][k]=num[i-1][j][k]+num[i-1][j-1][k]+num[i-1][j][k-1]-num[i-1][j-1][k-1];
}
}
}
while(scanf("%d%d%d",&a,&b,&c)==3)
{
if(a==-1&&b==-1&&c==-1)
break;
if(a<=0||b<=0||c<=0)
printf("w(%d, %d, %d) = 1\n",a,b,c);
else if(a>20||b>20||c>20)
printf("w(%d, %d, %d) = %d\n",a,b,c,num[20][20][20]);
else
printf("w(%d, %d, %d) = %d\n",a,b,c,num[a][b][c]);
}
return 0;
}