欧拉计划#15

昨晚我想解决这个问题 challenge #15 from Project Euler :

从a的左上角开始 回溯)到右下角 角落。

_{projecteuler.net )}

2020电网?

我觉得这应该不难,所以我写了一个基本的递归函数:

const int gridSize = 20;

// call with progress(0, 0)
static int progress(int x, int y)
{
    int i = 0;

    if (x < gridSize)
        i += progress(x + 1, y);
    if (y < gridSize)
        i += progress(x, y + 1);

    if (x == gridSize && y == gridSize)
        return 1;

    return i;
}

很明显,我走错了方向。你将如何解决这个问题?我认为应该用一个方程来解决,而不是像我这样的方法,但不幸的是,这不是我的优点。

更新 :

我现在有了一个工作版本。基本上,当nm块仍然需要遍历时,它会缓存以前获得的结果。以下是代码和一些注释:

// the size of our grid
static int gridSize = 20;

// the amount of paths available for a "NxM" block, e.g. "2x2" => 4
static Dictionary<string, long> pathsByBlock = new Dictionary<string, long>();

// calculate the surface of the block to the finish line
static long calcsurface(long x, long y)
{
    return (gridSize - x) * (gridSize - y);
}

// call using progress (0, 0)
static long progress(long x, long y)
{
    // first calculate the surface of the block remaining
    long surface = calcsurface(x, y);
    long i = 0;

    // zero surface means only 1 path remains
    // (we either go only right, or only down)
    if (surface == 0)
        return 1;

    // create a textual representation of the remaining
    // block, for use in the dictionary
    string block = (gridSize - x) + "x" + (gridSize - y);

    // if a same block has not been processed before
    if (!pathsByBlock.ContainsKey(block))
    {
        // calculate it in the right direction
        if (x < gridSize)
            i += progress(x + 1, y);
        // and in the down direction
        if (y < gridSize)
            i += progress(x, y + 1);

        // and cache the result!
        pathsByBlock[block] = i;
    }

    // self-explanatory :)
    return pathsByBlock[block];
}

将其调用20次,对于大小为11到2020的网格,将产生以下输出:

There are 2 paths in a 1 sized grid
0,0110006 seconds

There are 6 paths in a 2 sized grid
0,0030002 seconds

There are 20 paths in a 3 sized grid
0 seconds

There are 70 paths in a 4 sized grid
0 seconds

There are 252 paths in a 5 sized grid
0 seconds

There are 924 paths in a 6 sized grid
0 seconds

There are 3432 paths in a 7 sized grid
0 seconds

There are 12870 paths in a 8 sized grid
0,001 seconds

There are 48620 paths in a 9 sized grid
0,0010001 seconds

There are 184756 paths in a 10 sized grid
0,001 seconds

There are 705432 paths in a 11 sized grid
0 seconds

There are 2704156 paths in a 12 sized grid
0 seconds

There are 10400600 paths in a 13 sized grid
0,001 seconds

There are 40116600 paths in a 14 sized grid
0 seconds

There are 155117520 paths in a 15 sized grid
0 seconds

There are 601080390 paths in a 16 sized grid
0,0010001 seconds

There are 2333606220 paths in a 17 sized grid
0,001 seconds

There are 9075135300 paths in a 18 sized grid
0,001 seconds

There are 35345263800 paths in a 19 sized grid
0,001 seconds

There are 137846528820 paths in a 20 sized grid
0,0010001 seconds

0,0390022 seconds in total

我接受丹本的回答,因为他的回答最能帮助我找到这个解决方案。但投票也给了蒂姆·古德曼和阿戈斯:)

奖金更新 :

// the size of our grid
const int gridSize = 20;

// magic.
static Func<A1, A2, R> Memoize<A1, A2, R>(this Func<A1, A2, R> f)
{
    // Return a function which is f with caching.
    var dictionary = new Dictionary<string, R>();
    return (A1 a1, A2 a2) =>
    {
        R r;
        string key = a1 + "x" + a2;
        if (!dictionary.TryGetValue(key, out r))
        {
            // not in cache yet
            r = f(a1, a2);
            dictionary.Add(key, r);
        }
        return r;
    };
}

// calculate the surface of the block to the finish line
static long calcsurface(long x, long y)
{
    return (gridSize - x) * (gridSize - y);
}

// call using progress (0, 0)
static Func<long, long, long> progress = ((Func<long, long, long>)((long x, long y) =>
{
    // first calculate the surface of the block remaining
    long surface = calcsurface(x, y);
    long i = 0;

    // zero surface means only 1 path remains
    // (we either go only right, or only down)
    if (surface == 0)
        return 1;

    // calculate it in the right direction
    if (x < gridSize)
        i += progress(x + 1, y);
    // and in the down direction
    if (y < gridSize)
        i += progress(x, y + 1);

    // self-explanatory :)
    return i;
})).Memoize();

顺便说一下,我想不出更好的方法来使用这两个参数作为字典的键。我在谷歌上搜索了一下,似乎这是一个常见的解决方案。哦,好吧。