hdu 1398 Square Coins(母函数/完全背包)

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11418    Accepted Submission(s): 7827

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

 

Source

Asia 1999, Kyoto (Japan)

题意：求用硬币组成n有多少种方案，硬币必须是平方数（1,4,9,...289）  n不会超过300 故硬币最大面值是289了

思路：关于母函数可以看这位大神的博客，我也是刚在这里学习来的。

http://blog.csdn.net/vsooda/article/details/7975485

这道题其实是在整数分解的模板上面改一下得来，求解过程中的三重循环一定要理解理解再理解，具体可以看我代码和注释。

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;
int a[350],b[350];///a[i]数组表示当前拼出i的方案数，b数组也是一个意思，只不过a是最终结果，而b只保存当前这一次的结果
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=0; i<=n; i++)///这里表示的是用1元硬币（无限个）拼出0~n元的方案数， 显然都是1种
        {
            a[i]=1;
            b[i]=0;
        }
        for(int i=2; i*i<=n; i++)///用i*i元硬币
        {
            for(int j=0; j<=n; j++)///枚举已有的0~n元硬币
                for(int k=0; k+j<=n; k+=i*i)///使用0个，1个...x个i*i元硬币（x*i*i+j<=n），这里的思想跟背包一样= =
                    b[k+j]+=a[j];///把结果累加
            for(int j=0; j<=n; j++)///更新结果
            {
                a[j]=b[j];
                b[j]=0;///置零，为下一次计算做准备
            }
        }
        printf("%d\n",a[n]);
        /*int dp[350];
        memset(dp,0,sizeof(dp));
        for(int i=0;i<=n;i++)
            dp[i]=1;
        for(int i=2;i*i<=n;i++)
            for(int j=i*i;j<=n;j++)
                dp[j]+=dp[j-i*i];
        printf("%d\n",dp[n]);*/
    }
    return 0;
}