HDU-1331-Function Run Fun(动态规划3）

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

类似动态规划，避免重复计算，把先前算的拿数组存起来，后面直接那来用就可以了

#include <stdio.h>
#include <string.h>
int a,b,c;
int ans[21][21][21];
int w(int a,int b,int c)
{
    if(a<=0||b<=0||c<=0)
    {
        return 1;
    }
    if(a>20||b>20||c>20)
    {
        return w(20,20,20);
    }
    if(ans[a][b][c])
        return ans[a][b][c];
    if(a<b&&b<c)
    {
        ans[a][b][c]=w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c);
    }
    else
    {
        ans[a][b][c]=w(a-1,b,c) + w(a-1,b-1,c) + w(a-1,b,c-1) - w(a-1,b-1,c-1);
    }
    return ans[a][b][c];
}

int main()
{
    while(scanf("%d%d%d",&a,&b,&c)&&a+b+c!=-3)
    {
        memset(ans,0,sizeof(ans));
        w(a,b,c);
        printf("w(%d, %d, %d) = %d\n",a,b,c,w(a,b,c));
    }
    return 0;
}