hdu 1398 母函数 (三)

最新推荐文章于 2019-09-30 17:48:40 发布

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#combinations #integer #shapes #numbers #input #each

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本文探讨了一种名为SquareCoins的特殊货币系统，该系统仅使用平方数值的硬币，并提供了一个算法来计算支付特定金额的不同组合数量。通过示例说明了如何利用这些硬币进行支付，展示了算法的具体实现。

Square Coins

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4205 Accepted Submission(s): 2857

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

#include<iostream>
using namespace std;
int main(){
int 	c1[11111],c2[11111];
	int n;
	while(cin>>n){
		if(n==0)break;
		int i,j,k;
		for(i=0;i<=n;i++){
			c1[i]=1;
			c2[i]=0;
		}
		for(i=2;i<=17;i++){
			for(j=0;j<=n;j++){
				for(k=0;k+j<=n;k+=i*i){
					c2[j+k]+=c1[j];
				}
			}
		
		for(j=0;j<=n;j++){
			c1[j]=c2[j];
			c2[j]=0;
		}
		/*for(i=0;i<=n;i++){
			sum+=c1[i];
		}
		cout<<sum<<endl;
	}*/
		
		}
		cout<<c1[n]<<endl;
	}
return 0;
}