Function Run Fun
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
用純遞迴的思想會超時,用邊搜尋邊記憶的思路
#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) 快速冪求次方
{
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]) 這很重要,沒有這個,就是死迴圈
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); 邊搜尋,邊記憶
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)); 我把這小寫看成了大寫。
}
return 0;
}