hdu 1398 Square Coins

最新推荐文章于 2019-09-30 17:48:40 发布

转载最新推荐文章于 2019-09-30 17:48:40 发布 · 52 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/nefu929831238/p/6013888.html

本文介绍了一个关于整数拆分的问题变形，即如何使用特定面额的硬币来支付给定金额的不同组合数量。通过编程实现，探讨了递推算法的应用，并给出了一段C++代码示例。

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

1
4
27


解析：
    这是整数拆分的变形题，提供的整数变成平方项

#include <iostream>
#include <cstring>
using namespace std;

int main()
{
    int n;
    int c1[305],c2[305];
    while(cin>>n&&n)
    {
        for(int i=0;i<=n;i++)
        {
            c1[i]=1;
            c2[i]=0;
        }
        for(int i=2;i<=17;i++)
        {
            for(int j=0;j<=n;j++)
             for(int k=0;k+j<=n;k+=i*i)
              c2[k+j]+=c1[j];
            memcpy(c1,c2,sizeof(c2));
            memset(c2,0,sizeof(c2));
        }
        cout<<c1[n]<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/nefu929831238/p/6013888.html