UVa 11375 Matches (DP&高精度)

最新推荐文章于 2019-12-04 14:05:39 发布

最新推荐文章于 2019-12-04 14:05:39 发布 · 146 阅读

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本文介绍了一个使用动态规划解决火柴拼数字问题的算法，输入为火柴数量，输出为能拼出的不同非负整数的数量。算法通过预处理得到每个数字所需的火柴数，并使用动态规划计算总组合数。

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:

GivenNmatches, 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 integerN(1 ≤N≤ 2000).

Output

For eachN, output the number of different (non-negative) numbers representable if you haveNmatches.

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;
}