HDU - 1398 Square Coins (母函数)
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.
InputThe 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.
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.
OutputFor 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 0Sample Output
1 4 27
题意：给你1个n,你有 1 4 9 16 25 ..... i^2 等元素，每个有无限个，现在问你你有几种方案，由任意个元素组成n
比如
10 4种方案
1+9
1+1+....+1(10个)
4+1+.....+1(6个)
4+4+1+1
解：c1[i]=j代表的是组成i的方案有j种，c2的意义和c1相同，不过c2是再引入1个新的元素后，方案的变化，然后我们把c2的值赋给c1，c2重新归零
比如 10
我们首先明白可以组成他的元素为 1,4,9，因为每个元素有无限个，不过我们得知道c1[0]=1，因为1个都不选也是一种状态
我们先拿出 元素k= 1 ，显然 for( i ,0 , 10) c1[i] =1 唯一的方案就是 i 个1
然后拿出 元素 k=4 从 for(i,0,10)，每个i 都可以 一直加k 直到值大于 10为止
0 : c2[0]+=c1[0] c2[4]+=c1[0] c2[8]+=c1[0]
1: c2[1]+=c1[1] c2[5]+=c1[1] c2[9]+=c1[1]
2 : c2[2]+=c1[2] c2[6]+=c1[2] c2[10]+=c1[2]
............
然后我们把c2的值赋给c1,清空c2
拿出 元素 k=9 for（i，0,10） 每个i 都可以 一直加k 直到值大于 10为止
0：c2[0]+=c1[0] c2[9]+=c1[0]
............
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