Function Run Fun
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 13517 | Accepted: 7040 |
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.
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
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 5 using namespace std; 6 7 int dp[21][21][21]; 8 9 int calw(int a, int b, int c) 10 { 11 if(a <= 0 || b <= 0 || c <= 0) 12 { 13 //[a][b][c] = 1; //注意这里不能直接写,以为a、b、c可能超出数组范围 14 return 1; 15 } 16 if(a > 20 || b > 20 || c > 20) 17 { 18 dp[20][20][20] = calw(20, 20, 20); 19 return dp[20][20][20]; 20 } 21 if(dp[a][b][c]) 22 return dp[a][b][c]; 23 if(a < b && b < c) 24 { 25 dp[a][b][c] = calw(a, b, c - 1) + calw(a, b - 1, c - 1) - calw(a, b-1, c); 26 return dp[a][b][c]; 27 } 28 dp[a][b][c] = calw(a-1, b, c) + calw(a - 1, b - 1, c) + calw(a - 1, b, c - 1) - calw(a - 1, b - 1, c - 1); 29 return dp[a][b][c]; 30 } 31 32 int main() 33 { 34 int a, b, c; 35 while(cin >> a >> b >> c && a != -1 || b != -1 || c != -1) 36 { 37 printf("w(%d, %d, %d) = %d\n",a, b, c, calw(a, b, c)); 38 } 39 return 0; 40 }