HDU 1331 Function Run Fun

Function Run Fun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2173    Accepted Submission(s): 1104

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

Pacific Northwest 1999

 

Recommend

Ignatius.L

 

 

这道题要是直接打公式肯定会超时，所以采用记忆化，避免重复计算。

这道题和 HDU 1165不同，同样的是给出公式，那道题目就不能用记忆化，因为坐标不是递减的，容易超出坐标，并且那道题递归是嵌套的，不好用数组记录。

再看本题，本题递归中不含递归而且坐标是递减的。所以用记忆化的方法。

不同的题目处理方法不同

#include<cstdio>
#include<cmath>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int w[100][100][100];
int f(int a,int b,int c)
{
    if(a<=0||b<=0||c<=0)
        return 1;
    else if(a>20||b>20||c>20)
        return f(20,20,20);
    else if(a<b&&b<c)
    {
        if(!w[a][b][c])
            w[a][b][c]=f(a,b,c-1)+f(a,b-1,c-1)-f(a,b-1,c);
        return w[a][b][c];
    }
    else
    {
        if(!w[a][b][c])
            w[a][b][c]=f(a-1,b,c)+f(a-1,b-1,c)+f(a-1,b,c-1)-f(a-1,b-1,c-1);
        return w[a][b][c];
    }
}
int main()
{
    int a,b,c;
    memset(w,0,sizeof(w));
    while(scanf("%d %d %d",&a,&b,&c)!=EOF)
    {
        if(a==-1&&b==-1&&c==-1)
            break;
        printf("w(%d, %d, %d) = %d
",a,b,c,f(a,b,c));
    }
    return 0;
}