HDU 1331 Function Run Fun

水题一道，亏我还想了半天，之前我还找规律，找了半天还没找到，之后一看else
w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1)
顿时明白后面的都是前面得来的，
直接打表不就好了嘛~~~~（害我想辣么久）
Function Run Fun

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3704 Accepted Submission(s): 1805

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

代码：

#include<stdio.h>int aa[21][21][21];int f(int a,int b,int c){    if(a<=0||b<=0||c<=0)        return 1;    if(a<=b||a<=c)        return 1<<a;    return aa[a-1][b][c]+aa[a-1][b-1][c]+aa[a-1][b][c-1]-aa[a-1][b-1][c-1];}int main(){    int a,b,c,i,j,k;    for(i=0; i<=20; i++)        for(j=0; j<=20; j++)            for(k=0; k<=20; k++)                aa[i][j][k]=f(i,j,k);    while(scanf("%d%d%d",&a,&b,&c)&&(a!=-1||b!=-1||c!=-1))    {     printf("w(%d, %d, %d) = ",a,b,c);     if(a<=0||b<=0||c<=0)      printf("1\n");   else  if(a>20||b>20||c>20)      printf("%d\n",1<<20);     else      printf("%d\n",aa[a][b][c]);    }}