×joj 1175 The Binomial Function 递归，递归优化，非递归

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本文介绍了如何使用递归和备忘录优化技术来高效计算二项式函数值，适用于输入范围为0到30的整数对。包括三种递归方法：普通递归、带备忘录的递归和非递归版本，并通过实例展示了不同方法的实现和性能比较。

1175: The Binomial Function

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	2698	800	Standard

1st Jilin University ACM International Collegiate Programming Contest

In this problem, you are to write a program to make the value of the binomial function:

where n and m are integers.

Input Specification

The input consists of several pairs of n and m(0<=n, m<=30). The end of input is marked by n=m=0, which should not be processed.

Output Specification

For each pair of n and m, you should print a line like this: "Binom(n, m) = v", where n, m and v have to be replaced by the values of n, m and v.

Sample Input

Sample Output

Binom(5, 5) = 1
Binom(7, 0) = 1
Binom(5, 9) = 32

/*
	包含两个参数的递归
	三个版本的递归
	普通递归
	带有备忘录（或者剪枝）
	非递归
*/

#include <stdio.h>
int array[31][31];
/*
	超时
*/
int Binom(int n,int m)
{
	if(n == 0 || n == m || m == 0)
		return 1;
	else 
		return Binom(n-1,m) + Binom(n-1,m-1);
}
/*
	带有备忘录的递归，类似动态规划
*/
int binom(int i, int j){ 
	if(i==0||j==0||i==j)
		return 1;
	else if(array[i][j]) //加一个剪枝
		return array[i][j];
	else
		return array[i][j]=binom(i-1,j-1) + binom(i-1,j);
}
/*
	 非递归版
	 在JOJ上貌似这个版本不比那个剪枝的快
*/
int binomAd(int n,int m)
{	    
	int i;
    for ( i=0;i<=30;++i)
    {
        array[i][0]=1;
        array[i][i]=1;
        array[0][i]=1;
    }
    for ( i=1;i<=30;++i)
     for (int j=1;j<=30;++j)
     {
         if (j!=i)
		   array[i][j]=array[i-1][j]+array[i-1][j-1];
     }
	 return array[n][m];
}
int main()
{
	int n,m;
	while(scanf("%d%d",&n,&m),n !=0||m != 0)
	{
		int t = binomAd(n,m);
		printf("Binom(%d, %d) = %d\n",n,m,t);
	}
	return 0;
}