hdoj1331

Function Run Fun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1323    Accepted Submission(s): 672

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 12 2 210 4 650 50 50-1 7 18-1 -1 -1 

 

Sample Output

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

 

Source

Pacific Northwest 1999

 

Recommend

Ignatius.L

递归会超时。

注意到：

f 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)

都只与已计算出来的值有关，可以用数组f[a][b][c]表示w(a,b,c)的值，当计算需要前面用到的值时可以再次利用，避免了递归的重复计算。

#include<stdio.h>int f[25][25][25];void calculate(int a, int b, int c) {    if (a <= 0 || b <= 0 || c <= 0)        f[a][b][c] = 1;    else if (a < b && b < c)        f[a][b][c] = f[a][b][c - 1] + f[a][b - 1][c - 1] - f[a][b - 1][c];    else        f[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];}int main(void) {    int a, b, c;    for (int i = 0; i <= 20; i++)        for (int j = 0; j <= 20; j++)            for (int k = 0; k <= 20; k++)                calculate(i, j, k);    while (scanf("%d%d%d", &a, &b, &c), !(a == -1 && b == -1 && c == -1)) {        printf("w(%d, %d, %d) = ", a, b, c);        if (a <= 0 || b <= 0 || c <= 0)            printf("1");        else if (a > 20 || b > 20 || c > 20)            printf("%d", f[20][20][20]);        else            printf("%d", f[a][b][c]);        printf("\n");    }    return 0;}