Function Run Fun （记忆化搜索）

题目链接：https://cn.vjudge.net/contest/269853#problem

题目描述：

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

思路：

 这里主要是用的递归，但若是每次计算都要递归一次的话，最后会超时，所以用一个三维数组将结果先存起来，使用时直接调用结果即可，不需要一次次的重复递归运算。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1010;
int dp[25][25][25];
int fun(int a,int b,int c)
{
        if(a<=0||b<=0||c<=0)
            return 1;
        else if(a>20||b>20||c>20)
            return dp[20][20][20];
        else
            return dp[a][b][c];
}
void yu()//用预处理
{
    for(int i=1;i<=20;i++)
        for(int j=1;j<=20;j++)
          for(int k=1;k<=20;k++)
           {
             if(i<j&&j<k)
                dp[i][j][k]=fun(i,j,k-1)+fun(i,j-1,k-1)-fun(i,j-1,k);
                else
                    dp[i][j][k]=fun(i-1,j,k)+fun(i-1,j-1,k)+fun(i-1,j,k-1)-fun(i-1,j-1,k-1);

           }
}


int main()
{
    int a,b,c;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
        yu();
        if(a==-1&b==-1&&c==-1)break;
        if(a>20&&b>20&&c>20)
           printf("w(%d, %d, %d) = 1048576\n",a,b,c);
        else
       {
         int sum=fun(a,b,c);
         printf("w(%d, %d, %d) = %d\n",a,b,c,sum);
        }
    }
    return 0;
}