【HDU 1331 Function Run Fun】+ 记忆搜索

Function Run Fun

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3911 Accepted Submission(s): 1911

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 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

一道简单的记忆搜索题 ：

#include<cstdio>
int dp[25][25][25];
int sol(int a,int b,int c)
{
if(a <= 0 || b <= 0 || c <= 0)
return 1;
if(a > 20 || b > 20 || c >20)
return sol(20,20,20);
if(dp[a][b][c])
return dp[a][b][c];
if(a < b && b < c)
dp[a][b][c] = sol(a,b,c - 1) + sol(a,b - 1 ,c - 1) - sol(a,b - 1, c);
else
dp[a][b][c] = sol(a - 1, b,c) + sol(a - 1 , b - 1,c) + sol(a - 1,b,c - 1) - sol(a - 1,b - 1,c - 1);
return dp[a][b][c];
}

int main()
{
int a,b,c;
while(scanf("%d %d %d",&a,&b,&c)!=EOF){
if(a == -1 && b == -1 && c == -1) break;
printf("w(%d, %d, %d) = %d\n",a,b,c,sol(a,b,c));
}
return 0;
}