Combinatorics——HDUOJ 1398

原题：

• 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

解题思路：
1. 水题，母函数的基础练习题，需要注意，钱币每个coins种类的个数都是无限的。
2. 对于这种题型，预处理还是有必要的。
有关母函数的可以参考我的这篇：http://blog.csdn.net/Ming991301630/article/details/78322855
代码：

#include<stdio.h>#include <iostream>using namespace std;short coins[17] = { 1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289 };int c1[301], c2[301];int SUM[301];//母函数int CoinsKinds(short m){    short i, j, k;    for (i = 0; i <= m; i++)    {        c1[i] = 1;        c2[i] = 0;    }    for (i = 0; i <= 15; i++)    {        if (coins[i] > m)break;        for (j = 0; j <= m; j ++)            for (k = 0; k + j <= m; k += coins[i+1])            {                c2[j + k] += c1[j];            }        for (j = 0; j <= m; j++)        {            c1[j] = c2[j];            c2[j] = 0;        }    }    return c1[m];}int main(){    short i;    for (i = 1; i <= 300; i++)    {        SUM[i] = CoinsKinds(i);    }    int m;    while(scanf("%d",&m)&&m!=0)    {                       cout << SUM[m] << endl;    }}