[ACM] hdu 1398 Square Coins （母函数）

Square Coins

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7452 Accepted Submission(s): 5050

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

解题思路：

本题的母函数为 (1+x+x^2+x^3+x^4........) * (1+x^4+x^8+x^12+x^16....) *(1+x^9+x^18+x^27....) * ..............................*(1+x^289+x^(289*2)+..............

1分的硬币有多少个 4分的硬币有多少个 9分的硬币有多少个

这个多了一些限制条件，在母函数模板里改几个地方就可以了。重在理解。

代码：

#include <iostream>
using namespace std;
int c[302],temp[302],n;

int main()
{
    while(cin>>n&&n)
    {
        for(int i=0;i<=n;i++)
        {
            c[i]=1;
            temp[i]=0;
        }
        for(int i=2;i*i<=n;i++)//i是代表多少个式子相乘，这里i*i<=n的意思是比如要求 x^16次方，第一个式子是以1打头的（1+x+x^2+x^3...)，1*x^16,所有的式子中最后一次出现x^16次方的是(1+x^16+x^32..)这是第4个式子。
        {
            for(int j=0;j<=n;j++)
                for(int k=0;k+j<=n;k+=i*i)//注意 k+=i*i ,因为比如当i等于2时，代表第2个式子1+x^4+x^8+x^12+x^16..，注意指数。
                    temp[j+k]+=c[j];
            for(int i=0;i<=n;i++)
            {
                c[i]=temp[i];
                temp[i]=0;
            }
        }
        cout<<c[n]<<endl;
    }
    return 0;
}