杭电ACM 1398 Square Coins（母函数）_acm square coins-优快云博客

本文探讨了使用特定面值为平方数的SquareCoins硬币支付任意金额的方法数量。通过编程实现了一种算法来解决这一问题，并给出了示例输入输出及代码实现。

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Square Coins

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

题目大意：有多种硬币，而且每种硬币的面值都是平方数（最大是17^2=289），给出一个数，请输出用这些硬币组成这个数有多少种方案

解题思想：在母函数模板的基础上改两处赋值的地方就行了。

AC代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
int c1[400],c2[400],i,j,k,n;
while(scanf("%d",&n)&&(n!=0))
{
  for(i=0;i<=n;i++)              //给第一个表达式初始化
  {
   c1[i]=1;
   c2[i]=0;
  }
  for(i=2;i*i<=n;i++)            //因为硬币都是数的平方，所以判断条件是i*i<=n
  {
   for(j=0;j<=n;j++)
   {
    for(k=0;k+j<=n;k=k+i*i)         //注意k值的变化，k=k+i*i;因为硬币只有平方数，所以每次都要加平方数
    {
     c2[k+j]=c2[k+j]+c1[j];
    }
   }
   for(j=0;j<=n;j++)
   {
    c1[j]=c2[j];
    c2[j]=0;
   }
  }
  printf("%d\n",c1[n]);
}
return 0;
}