HDUOJ ------1398

时间:2022-05-05
本文章向大家介绍HDUOJ ------1398,主要内容包括http://acm.hdu.edu.cn/showproblem.php?pid=1398、Square Coins、基本概念、基础应用、原理机制和需要注意的事项等,并结合实例形式分析了其使用技巧,希望通过本文能帮助到大家理解应用这部分内容。

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

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6697    Accepted Submission(s): 4521

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

 母函数.......

 1 #include<iostream>
 2 using namespace std;
 3 int main()
 4 {
 5     int n,i,j,k;
 6     while(cin>>n,n)
 7     {
 8       int c1[301]={0},c2[301]={0};
 9        for(j=0;j<=n;j++)
10        {
11            c1[j]=1;
12        }
13        for(i=2;i*i<=n;i++)
14        {
15            for(j=0;j<=n;j++)
16            {
17                for(k=0;k+j<=n;k+=i*i)
18                {
19                   c2[k+j]+=c1[j]; 
20                }
21            }
22            for(j=0;j<=n;j++)
23            {
24                c1[j]=c2[j];
25                c2[j]=0;
26            }
27        }
28        cout<<c1[n]<<endl;
29     }
30     return 0;
31 }