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
题解: 母函数。
有这么几种面值的金币,分别是1,4,9,16 ……即1^2,2^2,3^2,4^2……每种面值的金币都有无数个,问:给定一个数,用这些金币,有多少种不同的方法能组成这个数。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#include <stack>
#define MAX 0x3f3f3f3f
using namespace std;
typedef long long ll;
long long a[1111];
int main()
{
int n;
while(cin >> n && n)
{
int a[333],b[333];
for(int i=0;i<=333;i++)
{
a[i]=1;
b[i]=0;
}
for(int i=2;i<=17;i++)
{
for(int j=0;j<=n;j++)
{
for(int k=0;k+j<=n;k+=i*i)
{
b[k+j]+=a[j];
}
}
for(int j=0;j<=n;j++)
{
a[j]=b[j];
b[j]=0;
}
}
cout << a[n]<<endl;
}
return 0;
}