解题心得：

1、其实此题有两种做法，动态规划，母函数。个人更喜欢使用动态规划来做，也可以直接套母函数的模板

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12191    Accepted Submission(s): 8352

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)

dp：

#include<bits/stdc++.h>

using namespace std;

int main()

{

    int va[18];

    for(int i=1;i<=17;i++)

    {

        va[i] = i*i;

    }

    int n;

    int d[310];

    for(int i=0;i<310;i++)

    {

        d[i] = 1;

    }

    for(int i=2;i<=17;i++)

    {

        for(int j=va[i];j<310;j++)

            d[j] += d[j-va[i]];

    }

    while(~scanf("%d",&n))

    {

        if(n == 0)

            break;

        printf("%d\n",d[n]);

    }

}

母函数：

#include<bits/stdc++.h>

using namespace std;

int main()

{

    int c1[310],c2[310];

    int n;

    while(scanf("%d",&n) && n)

    {

        for(int i=0;i<=n;i++)

        {

            c1[i] = 1;

            c2[i] = 0;

        }

        for(int i=2;i*i<=n;i++)

        {

            for(int j=0;j<=n;j++)

            {

                for(int k=0;k+j<=n;k+=i*i)

                    c2[k+j] += c1[j];

            }

            for(int k=0;k<=n;k++)

            {

                c1[k] = c2[k];

                c2[k] = 0;

            }

        }

        printf("%d\n",c1[n]);

    }

}