HDU1579(DP)

Function Run Fun

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1625    Accepted Submission(s): 888

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

 

运用数组保留运算结果，避免重复计算造成资源浪费从而超时

 

#include<iostream>
using namespace std;
#define MAX 25

int dp[MAX][MAX][MAX];

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