hdu 1331 Function Run Fun

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

 

这道题其实把思路都告诉你了。。。

而题目上说数据稍微大点的话会，程序运行会用很多的时间，所以这里就用一个三维dp数组来记忆结果，使程序大大地加快速度。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 26
int dp[N][N][N];
int dfs(int a,int b,int c)
{
    //if(dp[a][b][c]!=0) return dp[a][b][c];
    if(a<=0 || b<=0 || c<=0)
      return 1;
     if(a>20 || b>20 || c>20)
       return dfs(20,20,20);
     if(dp[a][b][c]!=0) return dp[a][b][c];
     if(a<b && b<c)
       return dp[a][b][c]=dfs(a,b,c-1)+dfs(a,b-1,c-1)-dfs(a,b-1,c);
     else 
      return dp[a][b][c]=dfs(a-1,b,c)+dfs(a-1,b-1,c)+dfs(a-1,b,c-1)-dfs(a-1,b-1,c-1);
}
int main()
{
    int a,b,c;
    while(scanf("%d%d%d",&a,&b,&c)==3)
    {
        if(a==-1 && b==-1 && c==-1)
          break;
        memset(dp,0,sizeof(dp));
        int ans=dfs(a,b,c);
        printf("w(%d, %d, %d) = %d
",a,b,c,ans);
    }
    return 0;
}