HDOJ--1398Square Coins！！母函数

原文链接

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

 

原文大意：

在Silverland的人使用的硬币是数字的倍数，从1的倍数到17的倍数，每种硬币的个数不限，请问你每种金额有多少种组合方法。

思路：使用母函数。

代码：

#include<stdio.h>#include<string.h>#define max 300int a[max+1];  //数组a代表最后的结果 int b[max+1];  //数组b代表中间值 int main(){int i,j,k,n;//每种硬币的面值 int SquareCoins[18]={0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289};memset(a,0,sizeof(a));memset(b,0,sizeof(b));a[0]=1;                   //0代表什么都不用，必有一种方法 for(i=1;i<=17;i++)        //母函数的第几项 {for(j=0;j*SquareCoins[i]<=max;j++)             //每个函数里面的每一项 {for(k=0;k+j*SquareCoins[i]<=max;k++)       //每种方法相乘 {b[k+j*SquareCoins[i]]+=a[k];}}for(j=0;j<=max;j++)   //将b的值给a {a[j]=b[j];b[j]=0;}} while(scanf("%d",&n)&&n){printf("%d\n",a[n]);     //将方法打印出来 }return 0;}