hdu1398(母函数)

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6739    Accepted Submission(s): 4548

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

   
   
   
   

    
    
    
    2
10
30
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
4
27
   
   
   
   

 

Source

Asia 1999, Kyoto (Japan)

 

Recommend

Ignatius.L

 

本题题意为给出数量足够多的面额为平方数（1-17）的硬币，求能给定价值的组合数。

本体是个很明显的母函数，母函数的原理就不在这里说了。具体请看我转载的那片大牛的文章。

 

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;

int n;
int value[20];
int a[300+10];
int b[300+10];

void Generating_function()
{
	int i,j,k;
	memset(a,0,sizeof(a));
	memset(b,0,sizeof(b));
	a[0]=1;
	for(i=1;i<=17;i++)
	{
		for(j=0;j<=300;j++)
		{
			for(k=0;k*value[i]+j<=300;k++)
				b[k*value[i]+j]+=a[j];
		}
		memcpy(a,b,sizeof(b));
		memset(b,0,sizeof(b));
	}
}

int main()
{
	int i;
	for(i=1;i<=17;i++)
		value[i]=i*i;
	Generating_function();		
	while(~scanf("%d",&n))
	{
		if(0==n)
			break;
		printf("%d\n",a[n]);
	}
	return 0;
}