HDU 1398 Square Coins （母函数）

creator平

于 2018-12-06 17:21:06 发布

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分类专栏：母函数

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本文介绍了一种使用特定面额硬币（1^2至17^2的平方数）来支付任意金额的算法。通过预计算不同面额硬币的组合方式，实现了对任意小于300的支付金额的有效计算。

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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

题意：只有1^2,2^2,3^2——17^2，面值的硬币，求n有多少种组合方法。

思路：与HDU - 1028相同

#include<algorithm>
#include<string.h>
#include<stdio.h>
using namespace std;
int p[20],A[1000],B[1000],n;
int main()
{
    for(int i=1;i<=17;i++) p[i]=i*i;
    for(int i=0;i<=300;i++) B[i]=1;
    for(int i=2;i<=17;i++)
    {
        for(int i=0;i<=300;i++)
            A[i]=B[i];
        for(int k=1;p[i]*k<=300;k++)
        {
            for(int j=300;j>=0;j--)
            {
                B[p[i]*k+j]+=A[j];
            }
        }
    }
    while(~scanf("%d",&n),n)
    {
        printf("%d\n",B[n]);
    }
}