HDOJ1398 Square Coins（母函数）

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6129    Accepted Submission(s): 4135

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)

 

Recommend

Ignatius.L

 

第一个用母函数写的题，纪念下：

/*
构造母函数如下：
G(x)=(1+x+x^2+x^3+...)*(1+x^4+x^8+x^12+...)*(1+x^9+x^18+...)...(1+x^17+x^34+...)
*/ 

#include <cstdio>
#include <iostream>

using namespace std;

int main()
{
    int sum;
    int c1[305], c2[305];
    while(scanf("%d", &sum), sum)
    {
        for(int i = 0; i <= sum; ++i)
        {
            c1[i] = 1;//初始化为第一个括号各项的系数，之后再依次与后边的合并更新 
            c2[i] = 0;
        }
        

        for(int i = 2; i <= 17; ++i)// 共有17个大括号相乘 ，直接从第二个括号开始合并 
        {
            for(int j = 0; j <= sum; ++j)// 每次都合并到第一个括号中，这里 j 代表第一个括号中的各项系数 
            {
                for(int k = 0; k+j <= sum; k += i*i) //虽然括号之间是相乘关系，但是指数之间是相加关系 
                {
                    c2[k+j] += c1[j]; // c2 数组可以理解为每次存放的中间结果,因为每次都是后边的括号与第一个括号可并，而后边的括号系数都为一，所以只有第一个括号中的系数对合并后相应的系数有贡献
                }
            }
            for(int j = 0; j <= sum; ++j)
            {
                c1[j] = c2[j];
                c2[j] = 0;       // 记得每次合并一个括号后要把   c2 清零 
            } 
        }
        printf("%d\n", c1[sum]);
    }
    return 0;
}