动态规划：Square Coins

问题描述：

People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (=17²), i.e., 1-credit coins, 4-credit coins, 9-credit coins, ..., and 289-credit coins, are available in Silverland.

There are four combinations of coins to pay ten credits:

 

ten 1-credit coins,
one 4-credit coin and six 1-credit coins,
two 4-credit coins and two 1-credit coins, and
one 9-credit coin and one 1-credit coin.

 

Your mission is to count the number of ways to pay a given amount using coins of Silverland.

Input

The input consists of lines each containing an integer meaning an amount to be paid, followed by a line containing a zero. You may assume that all the amounts are positive and less than 300.

Output

For each of the given amount, one line containing a single integer representing the number of combinations of coins should be output. No other characters should appear in the output.

Sample Input

2

10

30

0

Sample Output

1

4

27

 

问题分析：

二维数组w，其中w[i][j]表示：使用<=i并且>=1的整数序列{1...i}的变系数平方和等于j的多项式个数。

构造递推式：w[i][j] = sum { w[i-1][j-k*i*i] }，其中k=0,1,... j-k*i*i >= 0，这个式子的意思是，使用<=i-1并且>=1的整数序列{1...i-1}的变系数平方和等于j-k*i*i的多项式个数的和，这个和等同于，使用<=i并且>=1的整数序列{1...i}的变系数平方和等于j的多项式个数。因为整数序列{1...i-1}的变系数平方和多项式再加上k*i*i就是整数序列{1...i}的变系数平方和多项式。

 

package com.iteye.caoruntao.zoj;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/**
 * @author caoruntao
 *
 */
public class SquareCoins {

	private static int w[][] = new int[18][301];
	
	public static void compute(){
		for(int i=1; i<18; i++){
			for(int j=0; j<301; j++){
				if(i==1 || j==0){
					w[i][j] = 1;
				}
				else{
					w[i][j] = 0;
				}
			}
		}
		
		for(int i=2; i<18; i++){
			for(int j=1; j<301; j++){
				for(int k=0; j-k*i*i>=0; k++){
					w[i][j] += w[i-1][j-k*i*i];
				}
			}
		}
	}
	
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String str = "";
		int n;
		compute();
		try {
			while(!(str = br.readLine()).equalsIgnoreCase("0")){
				n = Integer.parseInt(str);
				System.out.println(w[17][n]);
			}
		} catch (Exception e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		}
	}

}