学步园

11137 - Ingenuous Cubrency

Time limit: 3.000 seconds

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 (= 21³), 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

给你一个n ， 问能将n分为多少个立方数之和。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>

using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1

#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 10000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
const int MOD = (int)1e6 + 7;
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pi;
///#pragma comment(linker, "/STACK:102400000,102400000")
LL dp[MAXN];
int main()
{
//    freopen("out.text" , "w" , stdout);
    dp[0] = 1LL;
    for(int j  = 1 ; j * j * j <= MAXN ; j++)for(int a = j * j * j ; a <= MAXN ; a++)dp[a] += dp[a - j * j * j];
    int n;
    while(cin >> n)
    {
        cout << dp[n] << endl;
    }
    return 0;
}