poj1579 Function Run Fun-优快云博客

本文链接：https://blog.youkuaiyun.com/CSUST_ACM/article/details/7952166

本文讨论了一个复杂的三参数递归函数的优化方法，包括打表法和规律法，旨在提高函数在中等数值输入下的计算效率。通过实例展示了如何避免直接递归带来的性能瓶颈，提供了一种有效解决此类问题的策略。

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Function Run Fun

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

方法一：打表法

思路：运用递归会超时，故而只要计算出前w[i][j][k](0<=k<=20&&0<=i<=20&&0<=j<=20)

#include<stdio.h>
#include<iostream>
using namespace std;
int w[21][21][21];
void build()
{
     int i,j,k;
     for(i=0;i<=20;i++)
     for(j=0;j<=20;j++)
     for(k=0;k<=20;k++)
     {if(i<=0||j<=0||k<=0)
       w[i][j][k]=1;
       else if(i<j&&j<k)
       w[i][j][k]=w[i][j][k-1]+w[i][j-1][k-1]-w[i][j-1][k];
       else
       w[i][j][k]=w[i-1][j][k]+w[i-1][j-1][k]+w[i-1][j][k-1]-w[i-1][j-1][k-1] ;
     }
     }
int main()
{
    build();
int a,b,c,s;
while(scanf("%d%d%d",&a,&b,&c)!=EOF&&!(a==-1&&b==-1&&c==-1))
{   if(a<=0||b<=0|c<=0)
  s=1;
   else if(a>20||b>20||c>20)
     s=w[20][20][20];
 else
    s=w[a][b][c]  ;                                                     
 printf("w(%d, %d, %d) = %d\n",a,b,c,s);
}
return 0;
    }
    
方法二：规律法
#include<stdio.h>
#include<math.h>
int w(int a,int b,int c)
{
if(a<=0 || b<=0 || c<=0)
return 1;
else if(a>20 || b>20 || c>20)
return 1048576;       //  只要有一个数大于20，则返回w(20,20,20)=1048576,即2的20次方
else if(b>=a || c>=a)     //只要后面两个数有一个大于a,则返回2的a次方(开始超时后，打表得出的规律。。。)
return (double)pow((double)2,a);     //注意pow函数的强制类型转换,一不小心就编译错误了
else if(a<b && b<c)
return w(a,b,c-1)+w(a,b-1,c)-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;
double ans;
while(scanf("%d%d%d",&a,&b,&c)!=EOF)
{
if(a==-1 && b==-1 && c==-1)
break;
else
{
printf("w(%d, %d, %d) = ",a,b,c);
ans=w(a,b,c);
}
printf("%.0lf\n",ans);
}
return 0;
}