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. For example:

1 1 1
2 2 2
10 4 6
50 50 50
-1 7 18
-1 -1 -1

Output

Print the value for w(a,b,c) for each triple, like this:

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

题目大意：递归在进行很多次的时候耗时非常巨大，需要想办法避免

分析：这是一道递归的题目，但正如题目所说，直接递归时间会很长。题目中三个整数的范围都不大，可使用记忆化搜索，也就是打表的方法。

建立一个三维数组把所有的结果存进去。

#include <stdio.h>
int f[21][21][21];//存储递归的中间结果
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(f[a][b][c]>0)
        return f[a][b][c];
    if(a<b&&b<c)
        return w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c);
    else
        return w(a-1,b,c)+w(a-1,b-1,c)+w(a-1,b,c-1)-w(a-1,b-1,c-1);
}
int main()
{
    int a,b,c;
    //打表
    int i,j,k;
    for(i=0;i<21;i++)
        for(j=0;j<21;j++)
        for(k=0;k<21;k++)
        f[i][j][k]=w(i,j,k);
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
        if(a==-1&&b==-1&&c==-1)
            break;
        printf("w(%d, %d, %d) =",a,b,c);
        printf(" %d\n",w(a,b,c));
    }
    return 0;
}