A - Square Coins （基础母函数）

题目描述：

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 (=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

题意：给出1 4 9.........17^2的硬币，给出总钱数，算出有几种组合方式；

分析：母函数，每次内函数递增改为+i*i即可

代码：

#include<stdio.h>
using namespace std;
typedef long long ll;
const int maxn=17;
const int A=305;
ll num1[A],num2[A];
int n;
void intt()
{
    num1[0]=1;
            for(int i=1;i<=maxn;i++)
            {
                for(int j=0;j<A;j++)
                {
                    for(int k=0;k<A-j;k+=i*i)//此处改变即可
                    {
                        num2[j+k]+=num1[j];
                    }
                }
                for(int j=0;j<A;j++)
                {
                    num1[j]=num2[j];
                    num2[j]=0;
                }
            }
            return ;
}
int main()
{
    intt();
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            break;
        else
        {
            printf("%lld\n",num1[n]);
        }
    }
    return 0;
}