UVa 11137 Ingenuous Cubrency

Problem I: Ingenuous Cubrency

People in Cubeland use cubic coins. Not only the unit of currency iscalled acube but also the coins are shaped like cubes andtheir values are cubes. Coins with values of all cubic numbers up to9261 (= 21³), i.e., coins with the denominations of 1, 8,27, ..., up to 9261cubes, are available inCubeland.

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

Input consists of lines each containing an integer amount tobe 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 asingle integer representing the number of ways to pay the given amountusing the coins available in Cubeland.

Sample input

10 21779999

Output for sample input

2322440022018293

题意：

求将 n 写成若干个正整数的立方之和有多少种 种方法。

分析：

完全背包，这里假设将 n 写成若干个正整数的立方之和有d[ n ] 种方法。

状态转移方程：d[ j ] = d[ j ] + d[ j - c[ i ] ]  (c[ i ] 为 i 的立方)，

要想求得d[ j ]，需要从若干个正整数的立方之和为 j + c[ i ] 再加上一个立方和为 c[ i ]

的数转移而来。

代码：

#include <iostream>#include <cstdio>using namespace std;const int MAXI = 21;        //22^3 > nconst int MAXN = 10000 + 5;int c[22];long long d[MAXN];int main(){    for(int i = 1; i <= MAXI; i++) c[i] = i * i * i;    d[0] = 1;    for(int i = 1; i <= MAXI; i++)        for(int j = c[i]; j <= MAXN; j++)                d[j] += d[j - c[i]];    int n;    while(~scanf("%d", &n))        cout << d[n] << endl;    return 0;}