Square Coins (hdu1398)

本文介绍了一种使用母函数的方法来解决硬币组合问题。具体地,针对Silverland的方形硬币,每枚硬币的价值为平方数,文章提供了一个算法实现,用于计算支付特定金额的不同硬币组合数量。

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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

题意:有17中硬币,分别是1-17的平方,然后问你300以内的数可有多少中组成方法

这是简单的母函数模板提目

#include <iostream> #include <cstring> #include <cstdio> using namespace std; int s[20],ans[305],tmp[305]; int mu() {     for(int i=1;i<=17;i++)     s[i]=i*i;     memset(ans,0,sizeof(ans));     memset(tmp,0,sizeof(tmp));     for(int i=0;i<=300;i++)     ans[i]=1;     for(int i=2;i<=17;i++)     {         for(int j=0;j<=300;j++)         for(int k=0;k<=300&&k*s[i]+j<=300;k++)         {             tmp[j+k*s[i]]+=ans[j];         }         for(int j=0;j<=300;j++)         {             ans[j]=tmp[j];             tmp[j]=0;         }     }     return 0; } int main() {     int n;     while(cin>>n)     {         if(n==0)         break;         mu();         cout<<ans[n]<<endl;     }     return 0; }

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