题解报告：hdu 1398 Square Coins（母函数或dp）

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

解题思路：多种硬币（有币值为1^2,2^2,...,17^2共17种硬币）的组合！解法和此篇博文类似：题解报告：hdu 1284 钱币兑换问题（简单数学orDP）

AC代码一之dp(0ms)：

 #include<bits/stdc++.h>
 using namespace std;
 int main(){
     int n,dp[]={};
     for(int i=;i<=;++i)//预处理打表
         for(int j=i*i;j<;++j)
             dp[j]+=dp[j-i*i];
     while(cin>>n&&n){cout<<dp[n]<<endl;}
     return ;
 }

AC代码二之母函数(15ms)：生成函数G(x)=(1+x^1+x^2+...)(1+x^4+x^8+x^12+...)(1+x^9+x^18+...)(...)(1+x^289)。

 #include<bits/stdc++.h>
 using namespace std;
 int n,c1[],c2[];
 void init(){
     memset(c1,,sizeof(c1));
     memset(c2,,sizeof(c2));
     c1[]=;//指数为0的系数为1
     for(int i=;i<=;++i){
         for(int j=;j<=;++j)
             for(int k=;k+j<=;k+=i*i)
                 c2[k+j]+=c1[j];
         for(int j=;j<=;++j)
             c1[j]=c2[j],c2[j]=;
     }
 }
 int main(){
     init();
     while(~scanf("%d",&n)&&n){printf("%d\n",c1[n]);}
     return ;
 }