POJ：1579 Function Run Fun（递归转换+打表）

Function Run Fun

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 18326 Accepted: 9365

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 12 2 210 4 650 50 50-1 7 18-1 -1 -1

Sample Output

w(1, 1, 1) = 2w(2, 2, 2) = 4w(10, 4, 6) = 523w(50, 50, 50) = 1048576w(-1, 7, 18) = 1

Source

Pacific Northwest 1999

题目大意：根据题目给的递归函数w（a，b，c），实现它，使时间变短。

解题思路：看到a、b、c中任意一个大于20，那么输出w（20，20，20），所以上界最多是20；再看每个状态方程都依赖于前一个状态，那么三重for循环打表即可。

代码如下：

#include <cstdio>int dp[21][21][21];void dabiao(){for(int i=0;i<21;i++)//打到20是因为一旦有一个超过20了，那么都是输出的dp[20][20][20] {for(int j=0;j<21;j++){for(int k=0;k<21;k++){if(i==0||j==0||k==0)//含0的都得是1 ，且下面的俩转移方程依赖于前一个i-1，j-1，k-1，所以一定要把这些情况写上去 {dp[i][j][k]=1;}else{if(i<j&&j<k)//依赖于题目的方程 {dp[i][j][k]=dp[i][j][k-1]+dp[i][j-1][k-1]-dp[i][j-1][k];}else////依赖于题目的方程  {dp[i][j][k]=dp[i-1][j][k]+dp[i-1][j-1][k]+dp[i-1][j][k-1]-dp[i-1][j-1][k-1];}}}}}}int main(){dabiao();int a,b,c;while(scanf("%d%d%d",&a,&b,&c)!=EOF){if(a==-1&&b==-1&&c==-1)//结束标志 break;if(a<=0||b<=0||c<=0)//一旦有一个小于0那么输出1 {printf("w(%d, %d, %d) = 1\n",a,b,c);}else{if(a>20||b>20||c>20)//一旦有一个超过20了，那么都是输出的dp[20][20][20] {printf("w(%d, %d, %d) = %d\n",a,b,c,dp[20][20][20]);}else//输出那个数字 {printf("w(%d, %d, %d) = %d\n",a,b,c,dp[a][b][c]);}}}return 0;}