UVa 10717 Mint (DFS枚举4个数的lcm)

最新推荐文章于 2015-04-29 10:03:57 发布

原创最新推荐文章于 2015-04-29 10:03:57 发布 · 1.2k 阅读

0 ·

CC 4.0 BY-SA版权

UVa 同时被 3 个专栏收录

407 篇文章

订阅专栏

acm之路--数学

262 篇文章

订阅专栏

acm之路--好题/陷阱

154 篇文章

订阅专栏

本文介绍了一种用于设计具有特定高度且使用不同类型的硬币作为桌腿的咖啡桌的算法。该算法通过深度优先搜索（DFS）的方式找到最接近目标高度的桌腿长度组合。

10717 - Mint

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=115&page=show_problem&problem=1658

The Royal Canadian Mint has commissioned a new series of designer coffee tables, with legs that are constructed from stacks of coins. Each table has four legs, each of which uses a different type of coin. For example, one leg might be a stack of quarters, another nickels, another loonies, and another twonies. Each leg must be exactly the same length.

Many coins are available for these tables, including foreign and special commemorative coins. Given an inventory of available coins and a desired table height, compute the lengths nearest to the desired height for which four legs of equal length may be constructed using a different coin for each leg.

Input consists of several test cases. Each case begins with an integers: 4 <= n <= 50 giving the number of types of coins available, and 1 <= t <= 10 giving the number of tables to be designed. n lines follow; each gives the thickness of a coin in hundredths of millimetres. t lines follow; each gives the height of a table to be designed (also in hundredths of millimetres). A line containing 0 0 follows the last test case.

For each table, output a line with two integers: the greatest leg length not exceeding the desired length, and the smallest leg length not less than the desired length.

Sample Input

Output for Sample Input

800 1200
2000 2000

用dfs更简单（懒得写4个for），注意枚举了4个数后把所有高度的桌腿都判断一遍，这样更快。

完整代码：

/*0.048s*/

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int n, t;
int a[55], h[55], maxh[55], minh[55];

int gcd(int a, int b)
{
	return b ? gcd(b, a % b) : a;
}

int lcm(int a, int b)
{
	return a / gcd(a, b) * b;
}

void dfs(int now, int cnt, int cur)
{
	if (cnt == 4)
	{
		for (int i = 0; i < t; i++)
		{
			int temp = h[i] / now * now;///小技巧
			if (temp == h[i])
				maxh[i] = minh[i] = h[i];
			else
			{
				if (maxh[i] == -1 || temp > maxh[i])
					maxh[i] = temp;
				temp += now;
				if (minh[i] == -1 || temp < minh[i])
					minh[i] = temp;
			}
		}
		return;
	}
	if (cur == n) return;
	dfs(now, cnt, cur + 1);
	int temp = lcm(now, a[cur]);
	dfs(temp, cnt + 1, cur + 1);
}

int main()
{
	int i;
	while (scanf("%d%d", &n, &t), n)
	{
		memset(maxh, -1, sizeof(maxh));
		memset(minh, -1, sizeof(minh));
		for (i = 0; i < n; ++i) scanf("%d", &a[i]);
		for (i = 0; i < t; ++i) scanf("%d", &h[i]);
		for (i = 0; i < n; ++i) dfs(a[i], 1, i + 1);///开始暴力！
		for (i = 0; i < t; ++i) printf("%d %d\n", maxh[i], minh[i]);
	}
	return 0;
}