Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20560		Accepted: 10325

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.

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.

Print the value for w(a,b,c) for each triple.

1 1 1

2 2 2

10 4 6

50 50 50

-1 7 18

-1 -1 -1

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

 /// POJ 1579 记忆化搜索入门

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <cstring>

 #include <cmath>

 #define ll long long int

 #define INF 0x3f3f3f3f

 using namespace std;

 const int MAXN = ;

 int d[MAXN][MAXN][MAXN];

 int dfs(int a, int b, int c)

 {

     if(a <=  || b <=  || c <= ) return ;

     if(a >  || b >  || c > ) return dfs(, , );

     if(d[a][b][c]) return d[a][b][c];

     if(a < b && b < c) d[a][b][c] = dfs(a, b, c-)+dfs(a, b-, c-) - dfs(a, b-, c);

     else d[a][b][c] = dfs(a-, b, c)+dfs(a-, b-, c)+dfs(a-, b, c-)-dfs(a-,b-,c-);

     return d[a][b][c];

 }

 int main()

 {

     int ans, A, B, C;

     memset(d, , sizeof(d));

     while(~scanf("%d%d%d", &A, &B, &C))

     {

         if(A == - && B == - && C == -) break;

         ans = dfs(A, B, C);

         printf("w(%d, %d, %d) = %d\n", A, B, C, ans);

     }

     return ;

 }