HDU1398(稍高级一些的母函数，对之前的补充）

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

#include<stdio.h>
int save[320],temp[320];
int a[30];
int main()
{
    int n;
    for(int i=0; i<=20; i++)
        a[i]=i*i;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=0; i<=n; i++)
        {
            save[i]=1;
            temp[i]=0;
        }
        for(int i=2; a[i]<=n; i++)
        {
            for(int j=0; j<=n; j++)
                for(int k=0; k+j<=n; k+=a[i])
                    temp[k+j]+=save[j];
            for(int k=0; k<=n; k++)
            {
                save[k]=temp[k];
                temp[k]=0;
            }
        }
        printf("%d
",save[n]);
    }
}

 只需要进行很少的修改就可以，但开始时由于我打了个a[i]的表，而且理解有问题出现了死循环的情况，以后的博客详细说（今天已经太晚了，对我来说）