hdoj 1331 Function Run Fun(模拟)

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Function Run Fun

http://acm.hdu.edu.cn/showproblem.php?pid=1331
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2587    Accepted Submission(s): 1256


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
 
//模拟

#include<cstdio>#include<cmath>#include<cstring>#include<cstdlib>#include<algorithm>using namespace std;int dp[25][25][25];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 w(20,20,20);      if(dp[a][b][c])        return dp[a][b][c];        if(a<b&&b<c)           return dp[a][b][c]=w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c);           else             return dp[a][b][c]=w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1);    }int main(){  int a,b,c;  while(~scanf("%d%d%d",&a,&b,&c))  {      if(a==-1&&b==-1&&c==-1)        break;        memset(dp,0,sizeof(dp));       printf("w(%d, %d, %d) = %d\n",a,b,c,w(a,b,c));  }  return  0;} 


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