uva 11137 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 1 cube 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

题目大意：有21种硬币（1~21的三次方），给出一个金额数，求用21种硬币可以组成该金额数的方法数。

解题思路：完全背包。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#define N 10005
using namespace std;
typedef long long ll;
ll coin[21] = {1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261};
ll dp[10005];
int main() {
    int n;
    while (scanf("%d", &n) != EOF) {
        memset(dp, 0, sizeof(dp));
        dp[0] = 1;
        for (int i = 0; i < 21; i++) {
            for (int j = coin[i]; j <= n; j++) {
                if (dp[j - coin[i]]) {
                    dp[j] += dp[j - coin[i]];
                }
            }
        }
        printf("%lld\n", dp[n]);
    }
    return 0;
}