Java e解释有助于未来读者更多地理解您的代码efficiently@icer这个问题的答案可能很好，很干净，至少更容易理解和解释。我要给他赏金。我想帮忙，但我绝对不觉得受到欢迎或感激。对不起，您可以删除此帖子。 #include <iostream>

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e解释有助于未来读者更多地理解您的代码efficiently@icer这个问题的答案可能很好，很干净，至少更容易理解和解释。我要给他赏金。我想帮忙，但我绝对不觉得受到欢迎或感激。对不起，您可以删除此帖子。

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

int mini(int n, int m) {
    return m * (m + 1) / 2;
}
int maxi(int n, int m) {
    return m * (2 * n - m + 1) / 2;
}

typedef std::vector<unsigned long long> Long1D;
typedef std::vector<Long1D> Long2D;
typedef std::vector<Long2D> Long3D;

int main(int argc, const char * argv[]) {
    int n, m, s;
    n = 45;
    m = 6;
    s = 21;

    if ((s < mini(n, m)) || (s > maxi(n, m))) {
        cout << 0 << endl;
        return 0;
    }

    Long3D dp(2, Long2D(m + 1, Long1D(s + 1)));

    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= min(i, m); ++j) {
            for (int k = 1; k <= s; ++k) {
                if ((k < mini(i, j)) || (k > maxi(i, j))) {
                    dp[i % 2][j][k] = 0;
                }
                else if ((k == mini(i, j)) || (k == maxi(i, j)) || j == 1) {
                    dp[i % 2][j][k] = 1;
                }
                else {
                    dp[i % 2][j][k] = 0;
                    // !IMPORTANT -- general situation: dp[i][j][k]=dp[i-1][j-1][k-j]+dp[i-1][j][k-j]
                    if (k - j > mini(i - 1, j - 1))
                        dp[i % 2][j][k] += dp[(i - 1) % 2][j - 1][k - j];
                    if (k - j < maxi(i - 1, j))
                        dp[i % 2][j][k] += dp[(i - 1) % 2][j][k - j];
                }
            }
        }
    }

    cout << dp[n % 2][m][s] << endl;
    return 0;
}

static void sumToValue(int limit, int sum, int count, List<Integer> resultIP) {
    if (limit >= 0 && sum == 0 && count == 0) {
        // print resultIP, because it is one of the answers.
        System.out.println("sum(" + Arrays.toString(resultIP.toArray()) + ")");
    } else if (limit <= 0 || count == 0 || sum <= 0) {
        // not what we want
        return;
    } else {
        // Two options: choose current limit number or not
        sumToValue(limit - 1, sum, count, resultIP);// Not choose the limit
                                                    // number

        // or choose the limit number
        List<Integer> resultNext = new ArrayList<Integer>(resultIP);// copy
                                                                    // resultIP
        resultNext.add(limit);
        sumToValue(limit - 1, sum - limit, count - 1, resultNext);
    }
}

static void sumToValueCount(int limit, int sum, int count) {
    int dp[][][] = new int[limit + 1][sum + 1][count + 1];
    for (int i = 0; i <= limit; i++) {
        for (int j = 0; j <= sum; j++) {
            for (int k = 0; k <= count; k++) {
                if (j == 0 && k == 0) {
                    dp[i][j][k] = 1;
                } else if (i == 0 || j <= 0 || k == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // check to prevent negative index
                    if (j - i >= 0) {
                        // two options: choose the number or not choose the number
                        dp[i][j][k] = dp[i - 1][j - i][k - 1] + dp[i - 1][j][k];
                    } else {
                        dp[i][j][k] = dp[i - 1][j][k];
                    }
                }
            }
        }
    }
    System.out.println(dp[limit][sum][count]);
}

//limit is 45, sum is the sum we want, count is 6 referring to the question.
sumToValue(45, 255, 6, new ArrayList<Integer>());
sumToValueCount(45, 255, 6);