Function Run Fun

Function Run Fun

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 25   Accepted Submission(s) : 17

Problem 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

PKU

 暴力枚举水题，每个值对应的数据都是固定的，只需要从最小值到最大值依次递归，开一个三维数组，把每一次的对应的数据保存到里面，然后根据输入的值输出即可。

#include<stdio.h>

int main()
{
    int a,b,c;
    int num[21][21][21];
    for(a=0;a<=20;a++)
        for(b=0;b<=20;b++)
            for(c=0;c<=20;c++)/*循环，依次按照题目要求，获取数据*/
            {
                  if(a<=0||b<=0||c<=0)num[a][b][c]=1;
                else if(a<b&&b<c)
                    num[a][b][c]=num[a][b][c-1]+num[a][b-1][c-1]-num[a][b-1][c];
                else
                    num[a][b][c]=num[a-1][b][c]+num[a-1][b-1][c]+num[a-1][b][c-1]-num[a-1][b-1][c-1];
             }
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)/*根据输入要求输出对应的值*/
    {
        if(a==-1&&b==-1&&c==-1)break;
        if(a<=0||b<=0||c<=0)
        printf("w(%d, %d, %d) = 1\n",a,b,c,num[a][b][c]);
        else if(a>20||b>20||c>20)
            printf("w(%d, %d, %d) = %d\n",a,b,c,num[20][20][20]);
        else
            printf("w(%d, %d, %d) = %d\n",a,b,c,num[a][b][c]);
    }
    return 0;
}