Function Run Fun（北大2015上机考试）_fuction run fun的c语言-优快云博客

本文介绍了一种递归函数的高效实现方法，并通过一个具体的三参数递归函数w(a,b,c)来展示如何避免重复计算，减少运行时间。文中提供了一个使用缓存技术的C语言实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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
Source

Pacific Northwest 1999

代码：

#include<stdio.h>
#include<string.h>
int buf[50][50][50];//用于保存被计算过的值
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(buf[a][b][c]){
return buf[a][b][c];//这句非常关键，否则会导致递归超时，把计算过的值缓存下来
}

if (a<b && b<c){
buf[a][b][c]=w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c);
}
else{
buf[a][b][c]=w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1);
}
return buf[a][b][c];
}
int main(){
int a,b,c;
memset(buf, 0, sizeof(buf));//将数组的所有元素重置
while(scanf("%d%d%d", &a, &b, &c)!=EOF){
if(a==-1 && b==-1 && c==-1){
break;
}
printf("w(%d, %d, %d) = %d\n", a, b, c, w(a, b, c));
}
return 0;
}