Square Coins+ 母函数

最新推荐文章于 2022-05-03 21:36:52 发布

原创最新推荐文章于 2022-05-03 21:36:52 发布 · 612 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#母函数 #acm #C #Square

母函数专栏收录该内容

11 篇文章

订阅专栏

本文探讨了在银岛使用正方形硬币支付特定金额的方法，重点介绍了如何使用不同面额的硬币组合来支付十元，包括了硬币价值为1到289的所有平方数情况，并详细说明了计算不同支付方式的数量。

Square Coins

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6231 Accepted Submission(s): 4196

Problem Description

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 (=17^2), 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

Sample Output

1
4
27
 
/*思路：母函数*/
#include<iostream>
#include<cmath>
using namespace std ;
#define MAX 302
int c1[MAX] ,c2[MAX] ;
int main(void)
{
	int n;
	while(cin >> n && n)
	{
		memset(c1 , 0 , sizeof(c1));
		memset(c2 , 0 , sizeof(c2));
		int i, j , k ;
		for( i = 0 ; i <= n ; i++)
			c1[i] = 1;
		for( i = 2 ; i <= 17 ; i++)
		{
			for( j = 0 ; j <= n ; j ++ )
				for( k = 0 ; k + j <= n ; k +=i*i)
					c2[k+j] += c1[j] ;

			for( j = 0 ; j <= n ; j++)
			{
				c1[j] = c2[j] ;
				c2[j] = 0;
			}
		}
		cout<<c1[n]<<endl;
	}
	return 0;
}