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.
输入样例
2
10
30
0输出样例
1
4
27
用数组保存可选种类的母函数
附ac代码
#include<bits/stdc++.h>
using namespace std;
int val[17]={1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289};
int c1[310],c2[310];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int n;
while(cin>>n&&n)
{
for(int i=0;i<=n;++i)
{
c1[i]=1;c2[i]=0;
}
for(int i=2;i<=17;++i)//枚举需要合并的次数
{
for(int j=0;j<=n;++j)//枚举当前括号前的总式各幂次
for(int k=0;k+j<=n;k+=val[i-1])//当前括号的次数
{
c2[j+k]+=c1[j];
}
for(int j=0;j<=n;++j)
c1[j]=c2[j],c2[j]=0;
}
cout<<c1[n]<<'\n';
}
return 0;
}