Function Run Fun (ZOJ - 1168)

题面

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.

输入

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.

输出

Print the value for w(a,b,c) for each triple.

样例输入

11 1 1
22 2 2
310 4 6
450 50 50
5-1 7 18
6-1 -1 -1

样例输出

1w(1, 1, 1) = 2
2w(2, 2, 2) = 4
3w(10, 4, 6) = 523
4w(50, 50, 50) = 1048576
5w(-1, 7, 18) = 1

提示

无

思路

代码

 1using namespace std;
 2typedef long long ll;
 3
 4int ans[21][21][21];
 5int a, b, c;
 6
 7int w(int a, int b, int c) {
 8    if (a<=0 || b<=0 || c<=0)
 9        return 1;
10
11    if (a>20 || b>20 || c>20)
12        return ans[20][20][20];
13
14    if (ans[a][b][c]>0)
15        return ans[a][b][c];
16
17    if (a<b && b<c)
18        return w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c);
19
20    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);
21}
22
23int main(void) {
24    for (int i=0; i<21; i++)
25        for (int j=0; j<21; j++)
26            for (int k=0; k<21; k++)
27                ans[i][j][k] = w(i, j, k);
28
29    while (scanf("%d%d%d", &a, &b, &c)==3) {
30        if (a==-1 && b==-1 && c==-1)
31            break;
32        printf("w(%d, %d, %d) = %d\n", a, b, c, w(a, b, c));
33    }
34    return 0;
35}