UVa 11375 Matches (DP&高精度)

最新推荐文章于 2019-12-04 14:05:39 发布

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本文解析了UVA 11375 Matches问题，这是一个经典的动态规划问题。文章提供了完整的代码实现，并解释了如何利用动态规划求解给定数量的火柴能构成的不同非负数的数量。

11375 - Matches

Time limit: 2.000 seconds

http://uva.onlinejudge.org/index.php?option=onlinejudge&page=show_problem&problem=2370

We can make digits with matches as shown below:

Given N matches, find the number of different numbers representable using the matches. We shall only make numbers greater than or equal to 0, so no negative signs should be used. For instance, if you have 3 matches, then you can only make the numbers 1 or 7. If you have 4 matches, then you can make the numbers 1, 4, 7 or 11. Note that leading zeros are not allowed (e.g. 001, 042, etc. are illegal). Numbers such as 0, 20, 101 etc. are permitted, though.

Input

Input contains no more than 100 lines. Each line contains one integer N (1 ≤ N ≤ 2000).

Output

For each N, output the number of different (non-negative) numbers representable if you have N matches.

Sample Input

3
4

Sample Output

2
4

简单DP。

完整代码：

/*0.076s*/

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 2001;
const int c[] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6}; ///  c[i]表示数字i需要的火柴数

struct node
{
	int p[500], len;

	node()
	{
		memset(p, 0, sizeof(p));
		len = 0;
	}

	node(int a)
	{
		p[0] = a;
		len = 1;
	}

	node operator + (const node& a) const
	{
		node b;
		b.len = max(len, a.len);
		for (int i = 0; i < b.len; i++)
		{
			b.p[i] += p[i] + a.p[i];
			b.p[i + 1] = b.p[i] / 10;
			b.p[i] %= 10;
		}
		if (b.p[b.len] > 0) b.len++;
		return b;
	}

	node operator += (const node& a)
	{
		*this = *this + a;
		return *this;
	}

	void out()
	{
		if (len == 0) putchar('0');
		else
		{
			for (int i = len - 1; i >= 0; i--)
				printf("%d", p[i]);
		}
		putchar(10);
	}
} d[maxn];

int main()
{
	int i, j, n;
	d[0] = node(1);
	for (i = 0; i < maxn; ++i)
		for (j = 0; j < 10; ++j)
			if (i + c[j] < maxn && (i || j))
				d[i + c[j]] += d[i];
	d[6] += node(1);
	for (i = 2; i < maxn; ++i) d[i] += d[i - 1];
	while (~scanf("%d", &n))
		d[n].out();
	return 0;
}