Function Run Fun
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 15623 | Accepted: 8080 |
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
思路：若直接用递归，肯定会超时，需要用三维数组进行记忆化，减少时间。
1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 using namespace std; 5 int dp[22][22][22]; 6 int DFS(int a,int b,int c) 7 { 8 if(dp[a][b][c]!=-1) 9 return dp[a][b][c]; 10 if(a<=0||b<=0||c<=0) 11 return 1; 12 else if(a>20||b>20||c>20) 13 return DFS(20,20,20); 14 else if(a<b&&b<c) 15 return DFS(a, b, c-1) + DFS(a, b-1, c-1) - DFS(a, b-1, c); 16 else 17 return DFS(a-1, b, c) + DFS(a-1, b-1, c) + DFS(a-1, b, c-1) - DFS(a-1, b-1, c-1); 18 } 19 int main() 20 { 21 int a,b,c,i,j,k; 22 memset(dp,-1,sizeof(dp)); 23 for(i=0; i<=20; i++) 24 for(j=0; j<=20; j++) 25 for(k=0; k<=20; k++) 26 dp[i][j][k]=DFS(i,j,k); 27 while(cin>>a>>b>>c) 28 { 29 if(a==-1&&b==-1&&c==-1) 30 break; 31 if(a<=0||b<=0||c<=0) 32 printf("w(%d, %d, %d) = 1 ",a,b,c); 33 else if(a>20||b>20||c>20) 34 printf("w(%d, %d, %d) = %d ",a,b,c,dp[20][20][20]); 35 else 36 printf("w(%d, %d, %d) = %d ",a,b,c,dp[a][b][c]); 37 } 38 return 0; 39 }