HDU-1331（DP）

Function Run Fun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3414    Accepted Submission(s): 1683

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

 

题意：有一个函数，按照上面规则，给出参数，求函数值。

思路：DP。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int dp[22][22][22];
int a, b, c;

void calcu(){//打表，计算每一个状态的值
    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)
                    dp[i][j][k] = 1;
                else if(i < j && j < k)
                    dp[i][j][k] = dp[i][j][k-1] + dp[i][j-1][k-1] -
                                  dp[i][j-1][k];
                else
                    dp[i][j][k] = dp[i-1][j][k] + dp[i-1][j-1][k] +
                                  dp[i-1][j][k-1] - dp[i-1][j-1][k-1];
            }
        }
    }

}

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 dp[20][20][20];
    return dp[a][b][c];
}

int main()
{
    calcu();
    while(~scanf("%d%d%d", &a, &b, &c)){
        if(a==-1&&b==-1&&c==-1) break;
        printf("w(%d, %d, %d) = %d\n", a, b, c, w(a, b, c));
    }
    return 0;
}