Uva 11137 Ingenuous Cubrency 解题报告（递推）

Problem I: Ingenuous Cubrency

People in Cubeland use cubic coins. Not only the unit of currency is called a  cube but also the coins are shaped like cubes and their values are cubes. Coins with values of all cubic numbers up to 9261 (= 21 ³ ), i.e., coins with the denominations of 1, 8, 27, ..., up to 9261  cubes , are available in Cubeland.

Your task is to count the number of ways to pay a given amount using cubic coins of Cubeland. For example, there are 3 ways to pay 21 cubes: twenty one 1cube coins, or one 8 cube coin and thirteen 1 cube coins, or two 8 cube coin and five 1 cube coins.

Input consists of lines each containing an integer amount to be paid. You may assume that all the amounts are positive and less than 10000.

For each of the given amounts to be paid output one line containing a single integer representing the number of ways to pay the given amount using the coins available in Cubeland.

Sample input

10 
21
77
9999

Output for sample input

2
3
22
440022018293

    解题报告： 和上一篇博客 火柴 那题相似。问一个数可以有几种三次方和的拼法。仍然是递推，和上一题不同的是，这个数字组合是无序的，所以不能使用上一种线性的递推法。9999对应的答案提示我们用long long就可以了。

    笔者代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long LL;
const int maxn = 10000;
LL dp[22][maxn];

int main()
{
    dp[0][0] = 1;

    for(int i=1;i<22;i++)
    {
        for(int j=0;j<maxn;j++)
        {
            for(int a=0;j+a*i*i*i<maxn;a++)
            {
                dp[i][j+a*i*i*i]+=dp[i-1][j];
            }
        }
    }

    int n;
    while(~scanf("%d", &n))
        printf("%lld\n", dp[21][n]);
}

    需要这么复杂吗？哈哈，简化版：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long LL;
const int maxn = 10000;
LL d[maxn];

int main()
{
    d[0] = 1;
    for(int i=1;i<22;i++)
        for(int tmp=i*i*i,j=tmp;j<maxn;j++)
            d[j] += d[j-tmp];

    int n;
    while(~scanf("%d", &n))
        printf("%lld\n", d[n]);
}

    时间和空间都有优化了。