D - Function Run Fun

男娘i 2024-02-18 23:19 59阅读 0赞

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

用纯递归的思想会超时，用边搜索边记忆的思路

\#include<iostream>  
\#include<cstring>  
\#include<algorithm>  
\#include<cmath>  
\#define LL long long  
using namespace std;  
int aa\[21\]=\{1,2,4,\},t=0;  
int bb\[21\]\[21\]\[21\];  
int quick\_mod(int n,int k) 快速幂求次方  
\{  
int sum=1;  
n%=k;  
while(k!=0)  
\{  
if(k&1)  
\{  
sum=sum\*n;  
\}  
n=n\*n;  
k>>=1;  
\}  
return sum;  
\}  
int dj(int a,int b,int c)  
\{  
if(a<=0||b<=0||c<=0)  
\{  
return 1;  
\}  
if(a>20||b>20||c>20)  
return dj(20,20,20);  
if(bb\[a\]\[b\]\[c\]) 这很重要，没有这个，就是死循环  
return bb\[a\]\[b\]\[c\];  
if(a<b&&b<c)  
bb\[a\]\[b\]\[c\]=dj(a,b,c-1)+dj(a,b-1,c-1)-dj(a,b-1,c);  
else  
bb\[a\]\[b\]\[c\]=dj(a-1,b,c)+dj(a-1,b-1,c)+dj(a-1,b,c-1)-dj(a-1,b-1,c-1); 边搜索，边记忆  
return bb\[a\]\[b\]\[c\];  
\}  
int main()  
\{  
int i,j,k,a,b,c;  
memset(bb,0,sizeof(bb));  
for(i=3;i<21;i++)  
\{  
aa\[i\]=quick\_mod(2,i);  
\}  
  
while(~scanf("%d%d%d",&a,&b,&c))  
\{  
if(a==-1&&b==-1&&c==-1)  
break;  
if(a==b&&b==c&&a<20)  
\{  
printf("w(%d, %d, %d) = %d\\n",a,b,c,aa\[a\]);  
continue;  
\}  
if(a==b&&b==c&&a>20)  
\{  
printf("w(%d, %d, %d) = %d\\n",a,b,c,aa\[20\]);  
continue;  
\}  
dj(a,b,c);  
printf("w(%d, %d, %d) = %d\\n",a,b,c,dj(a,b,c)); 我把这小写看成了大写。  
\}  
return 0;  
\}