HDOJ Square Coins

最新推荐文章于 2017-11-16 20:38:40 发布

DeDeWo

最新推荐文章于 2017-11-16 20:38:40 发布

阅读量588

点赞数

CC 4.0 BY-SA版权

分类专栏：母函数文章标签： combinations integer shapes numbers input each

本文链接：https://blog.youkuaiyun.com/aihahaheihei/article/details/6783713

母函数专栏收录该内容

4 篇文章

订阅专栏

本文深入探讨了使用平方金币在Silverland中支付特定金额的方法，通过构造母函数解决组合问题，并提供了Java实现代码。针对不同金额，文章详细分析了支付方案的数量，展示了数学模型在实际问题解决中的应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Square Coins

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

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3774 Accepted Submission(s): 2590

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

Sample Output

这里的增量是i*i.

构造母函数: f(x)=(1+x+x^2+x^3+…)(1+x^4+x^4+x^12+…)(1+x^9+x^18+x^27+…)……

#include<iostream>
using namespace std;
int main(){
    int n;
    int c[301],temp[301];
    while(scanf("%d",&n),n){
          for(int i=0;i<=n;i++){
                c[i]=1;
                temp[i]=0;
          }
          for(int i=2;i<=17;i++){ //对每个表达式进行处理
                for(int j=0;j<=n;j++)
                      for(int k=0;j+k<=n;k+=i*i ) //增量是i*i
                             temp[j+k]+=c[j];
                for(int j=0;j<=n;j++){
                      c[j]=temp[j];
                      temp[j]=0;
                }
          }
          printf("%d\n",c[n]);
    }
    return 0;
}