Function Run Fun
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2173 Accepted Submission(s): 1104
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
Source
Recommend
Ignatius.L
这道题要是直接打公式肯定会超时,所以采用记忆化,避免重复计算。
这道题和 HDU 1165不同,同样的是给出公式,那道题目就不能用记忆化,因为坐标不是递减的,容易超出坐标,并且那道题递归是嵌套的,不好用数组记录。
再看本题,本题递归中不含递归而且坐标是递减的。所以用记忆化的方法。
不同的题目处理方法不同
1 #include<cstdio> 2 #include<cmath> 3 #include<cstring> 4 #include<stdlib.h> 5 #include<algorithm> 6 using namespace std; 7 int w[100][100][100]; 8 int f(int a,int b,int c) 9 { 10 if(a<=0||b<=0||c<=0) 11 return 1; 12 else if(a>20||b>20||c>20) 13 return f(20,20,20); 14 else if(a<b&&b<c) 15 { 16 if(!w[a][b][c]) 17 w[a][b][c]=f(a,b,c-1)+f(a,b-1,c-1)-f(a,b-1,c); 18 return w[a][b][c]; 19 } 20 else 21 { 22 if(!w[a][b][c]) 23 w[a][b][c]=f(a-1,b,c)+f(a-1,b-1,c)+f(a-1,b,c-1)-f(a-1,b-1,c-1); 24 return w[a][b][c]; 25 } 26 } 27 int main() 28 { 29 int a,b,c; 30 memset(w,0,sizeof(w)); 31 while(scanf("%d %d %d",&a,&b,&c)!=EOF) 32 { 33 if(a==-1&&b==-1&&c==-1) 34 break; 35 printf("w(%d, %d, %d) = %d\n",a,b,c,f(a,b,c)); 36 } 37 return 0; 38 }
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