Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17843		Accepted: 9112

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.

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.

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

1 1 1

2 2 2

10 4 6

50 50 50

-1 7 18

-1 -1 -1

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

记忆化搜索。照着思路走。不过把已经搜索过的就不要重复计算了

#include<iostream>

#include<cstdio>

#include<algorithm>

#include <string.h>

#include <math.h>

using namespace std;

int f[][][];

int dfs(int a,int b,int c){

    if(a<=||b<=||c<=) return ;

    if(a>||b>||c>) return dfs(,,);

    if(f[a][b][c]!=-) return f[a][b][c];

    if(a<b&&b<c) f[a][b][c] = dfs(a,b,c-)+dfs(a,b-,c-)-

        dfs(a,b-,c);

    else f[a][b][c]= dfs(a-,b,c)+dfs(a-,b-,c)+dfs(a-,b,c-)

        -dfs(a-,b-,c-);

    return f[a][b][c];

}

int main()

{

    int a,b,c;

    while(scanf("%d%d%d",&a,&b,&c)!=EOF){

        if(a==-&&b==-&&c==-) break;

        memset(f,-,sizeof(f));

        int ans = dfs(a,b,c);

        printf("w(%d, %d, %d) = %d\n",a,b,c,ans);

    }

    return ;

}