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
这道题给了dp的动态式子，做一个听话的好宝宝就好了。
#include<stdio.h>
int a,b,c,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);
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()
{
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;
}
