我必须创建一个接受两个整数的方法，让它们成为 n和 m , 并返回求和的方式数 m得到正数 n .
例如，像这样的方法调用 partition(6, 2)应该返回 3，因为有 3 种可能的方式。他们是 5 + 1 , 4 + 2 , 和 3 + 3 .顺便说一句，4 + 2与 2 + 4 相同，因此该方法不应将它们视为两个不同的变体。
有人知道问题的解决方案吗？

更新:n和 m不大于 150。

最佳答案

递归算法

计算整数 n 的所有分区与 m部分，递归算法是显而易见的选择。案例n, m , 算法遍历每一个选项 k = 1, 2, 3...对于第一部分，对于这些选项中的每一个，它都会以案例 n - k, m - 1 递归.例如:

n = 16, m = 4  
first part = 1  =>  recurse with n = 15, m = 3  
first part = 2  =>  recurse with n = 14, m = 3  
first part = 3  =>  recurse with n = 13, m = 3  
etc...

first part = 1  =>  second part = n - 1  
first part = 2  =>  second part = n - 2
first part = 3  =>  second part = n - 3
etc...

n = 9, m = 3  
partitions: 1+1+7   1+2+6   1+3+5   1+4+4  
            2+2+5   2+3+4  
            3+3+3  

n = 9, m = 3  
k = 1  =>  recurse with n = 8, m = 2, k >= 1  =>  1+7  2+6  3+5  4+4  
k = 2  =>  recurse with n = 7, m = 2, k >= 2  =>  2+5  3+4  
k = 3  =>  recurse with n = 6, m = 2, k >= 3  =>  3+3

n = 9, m = 3  
k = 1  =>  recurse with n = 8, m = 2, k >= 1  =>  1+7  2+6  3+5  4+4    
k = 2  =>  recurse with n = 5, m = 2, k >= 1  =>  1+4  2+3
k = 3  =>  recurse with n = 2, m = 2, k >= 1  =>  1+1

function partition(n, m) {
    if (m < 2) return m;
    if (n < m) return 0;
    if (n <= m + 1) return 1;
    var max = Math.floor(n / m);
    if (m == 2) return max;
    var count = 0;
    for (; max--; n -= m) count += partition(n - 1, m - 1);
    return count;
}

document.write(partition(6, 1) + "<br>");    // 1
document.write(partition(6, 2) + "<br>");    // 3
document.write(partition(9, 3) + "<br>");    // 7
document.write(partition(16, 4) + "<br>");   // 34
document.write(partition(150, 75) + "<br>"); // 8,118,264
// document.write(partition(150, 23));       // 1,901,740,434 (maximum for 150, takes > 10s)

function partition(n, m) {
    if (m < 2) return m;
    if (n < m) return 0;
    var memo = [];
    for (var i = 0; i < n - 1; i++) memo[i] = [];
    return p(n, m);

    function p(n, m) {
        if (n <= m + 1) return 1;
        if (memo[n - 2][m - 2]) return memo[n - 2][m - 2];
        var max = Math.floor(n / m);
        if (m == 2) return max;
        var count = 0;
        for (; max--; n -= m) count += (memo[n - 3][m - 3] = p(n - 1, m - 1));
        return count;
    }
}

document.write(partition(150, 23) + "<br>");  // 1,901,740,434
// document.write(partition(1000, 81));       // 4.01779428811641e+29

function partition(n, m) {
    if (m < 2) return m;
    if (n < m) return 0;
    var memo = [];
    for (var i = 0; i <= n - m; i++) memo[i] = [];
    return p(n, m);

    function p(n, m) {
        if (n <= m + 1) return 1;
        if (memo[n - m][m - 2]) return memo[n - m][m - 2];
        var max = Math.floor(n / m);
        if (m == 2) return max;
        var count = 0;
        for (; max--; n -= m) count += (memo[n - m][m - 3] = p(n - 1, m - 1));
        return count;
    }
}

document.write(partition(150, 23) + "<br>");  // 1,901,740,434
document.write(partition(150, 75) + "<br>");  // 8,118,264
document.write(partition(150, 127) + "<br>"); // 1255

关于java - 是否有一种有效的算法用于具有有限数量的部分的整数分区？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/32907406/