【HDU - 1398】Square Coins（生成函数）

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于 2019-04-12 10:49:03 发布

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分类专栏：生成函数 hdu 文章标签：生成函数 hdu

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探讨在银币国使用特殊方形硬币支付时，如何计算支付特定金额的不同组合方式。介绍了通过生成函数的方法来解决这一问题，并提供了一个算法实现的示例，该算法考虑了所有可能的硬币面值，直至289信用点。

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：

生成函数，稍微改一下模板就可以

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<string>
#include<vector>
using namespace std;
typedef  long long ll;
const ll MAX=2200;
int f[MAX];
ll a[MAX],b[MAX];
int main() {
	for(int i=1;i<=17;i++)
	{
		f[i]=i*i;
	}
	int n;
	while(cin>>n,n)
	{
		for(int i=0;i<=n;i++)
		{
			a[i]=1;
			b[i]=0;
		}
		for(int i=2;i<=17;i++)//总共有17种可能的硬币所以循环到17 
		{
			for(int j=0;j<=n;j++)
			{
				for(int k=0;j+k<=n;k+=f[i])//这里是枚举当前表达式指数的可能值，指数是f[i]的倍数，所以每次要加f[i] 
				{
					b[j+k]+=a[j];
				}
			}
			for(int j=0;j<=n;j++)
			{
				a[j]=b[j];
				b[j]=0;	
			}
		}
		cout<<a[n]<<endl;
	}
	return 0;
}