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
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
这道题其实把思路都告诉你了。。。
而题目上说数据稍微大点的话会，程序运行会用很多的时间，所以这里就用一个三维dp数组来记忆结果，使程序大大地加快速度。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 26
int dp[N][N][N];
int dfs(int a,int b,int c)
{
//if(dp[a][b][c]!=0) return dp[a][b][c];
if(a<= || b<= || c<=)
return ;
if(a> || b> || c>)
return dfs(,,);
if(dp[a][b][c]!=) return dp[a][b][c];
if(a<b && b<c)
return dp[a][b][c]=dfs(a,b,c-)+dfs(a,b-,c-)-dfs(a,b-,c);
else
return dp[a][b][c]=dfs(a-,b,c)+dfs(a-,b-,c)+dfs(a-,b,c-)-dfs(a-,b-,c-);
}
int main()
{
int a,b,c;
while(scanf("%d%d%d",&a,&b,&c)==)
{
if(a==- && b==- && c==-)
break;
memset(dp,,sizeof(dp));
int ans=dfs(a,b,c);
printf("w(%d, %d, %d) = %d\n",a,b,c,ans);
}
return ;
}