（POJ - 1579）Function Run Fun

最新推荐文章于 2021-08-01 12:05:02 发布

Hold_My_Own

最新推荐文章于 2021-08-01 12:05:02 发布

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分类专栏：递推文章标签： poj

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本文介绍了一个递归函数w(a,b,c)的优化实现，通过预计算并存储中间结果来避免重复计算，显著提高了效率。针对直接递归实现运行缓慢的问题，采用递推方法进行改进，并提供了一个C++实现示例。

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（POJ - 1579）Function Run Fun

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 19665 Accepted: 9920

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

思路：因为递归太慢，所以要转化成递推形式。注意输出格式即可。

#include<cstdio>
using namespace std;

const int maxn=25;
int w[maxn][maxn][maxn];

int main()
{
    for(int i=0;i<=20;i++)
        for(int j=0;j<=20;j++)
            for(int 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 a,b,c;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF&&(a!=-1||b!=-1||c!=-1))
    {
        if(a<=0||b<=0||c<=0) printf("w(%d, %d, %d) = 1\n",a,b,c);
        else if(a>20||b>20||c>20) printf("w(%d, %d, %d) = %d\n",a,b,c,w[20][20][20]);
        else printf("w(%d, %d, %d) = %d\n",a,b,c,w[a][b][c]);
    }   
    return 0;
}