Ingenuous Cubrency
People in Cubeland use cubic coins. Not only the unit of currency is called acube 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 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
题意：
输入正整数n(n <= 10000)，求将n写成若干个正整数的立方和有多少种方式。
分析：
建立多段图。结点(i,j)表示“使用不超过i的正整数的立方，累加和为j”这个状态，设dp(i,j)为从(0,0)到(i,j)的路径的条数，最终答案为dp(21,n)(22*22*22>n)。
转移方程是dp[i][j + a*i*i*i] += dp[i-1][j] (i + a*i*i*i<=n)
1 #include <cstdio> 2 #include <cstring> 3 const int maxn = 10000; 4 int n; 5 long long dp[22][maxn + 1]; 6 int main(){ 7 dp[0][0] = 1; 8 for(int i = 1 ; i <= 21 ; i++)for(int j = 0 ; j <= maxn ; j++) 9 for(int a = 0 ; a * i * i * i + j <= maxn ; a++) 10 dp[i][j + a * i * i * i] += dp[i - 1][j]; 11 while(~scanf("%d",&n)) printf("%lld ", dp[21][n]); 12 return 0; 13 }