【HDU 1398】【母函数入门题】Square Coins【硬币种类有1^2,2^2,3^2,4^2...17^2，这几种；输入n；求出能够组合成n的组合有多少种】

传送门：http://acm.split.hdu.edu.cn/showproblem.php?pid=1398

描述：

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11451    Accepted Submission(s): 7841

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

210300

 

Sample Output

1427

 

Source

Asia 1999, Kyoto (Japan)

 

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题意：

硬币种类有1^2,2^2,3^2,4^2...17^2，这几种；输入n；求出能够组合成n的组合有多少种；

思路：

基本的母函数题，与hdu1028相似，只是将原来的1，2，3，4，……换成了1^2,2^2,3^2,4^2...…

代码：

#include <bits/stdc++.h>using  namespace  std;#define rep(i,k,n) for(int i=k;i<=n;i++)const int N = 310;int num1[N], num2[N];void init(){  rep(i, 0, N - 1){    num1[i] = 1;    num2[i] = 0;  }  rep(i, 2, 17){    rep(j, 0, N - 1)//枚举已知范围 （前面所有项结果项结果）        for(int k = 0; k + j < N; k += i * i){ //枚举新增范围（下一项 ）        num2[k + j] += num1[j];      }    rep(j, 0, N - 1){//重置       num1[j] = num2[j];      num2[j] = 0;    }  }}int  main(){  int n;  init();  while(~scanf("%d", &n), n){    printf("%d\n", num1[n]);  }  return 0;}