UVA 165 Stamps （DFS深搜回溯）

 

Stamps

 

The government of Nova Mareterrania requires that various legal documents have stamps attached to them so that the government can derive revenue from them. In terms of recent legislation, each class of document is limited in the number of stamps that may be attached to it. The government wishes to know how many different stamps, and of what values, they need to print to allow the widest choice of values to be made up under these conditions. Stamps are always valued in units of $1.

 

This has been analysed by government mathematicians who have derived a formula for n(h,k), where h is the number of stamps that may be attached to a document, k is the number of denominations of stamps available, and n is the largest attainable value in a continuous sequence starting from $1. For instance, if h=3, k=2 and the denominations are $1 and $4, we can make all the values from $1 to $6 (as well as $8, $9 and $12). However with the same values of h and k, but using $1 and $3 stamps we can make all the values from $1 to $7 (as well as $9). This is maximal, so n(3,2) = 7.

 

Unfortunately the formula relating n(h,k) to h, k and the values of the stamps has been lost--it was published in one of the government reports but no-one can remember which one, and of the three researchers who started to search for the formula, two died of boredom and the third took a job as a lighthouse keeper because it provided more social stimulation.

The task has now been passed on to you. You doubt the existence of a formula in the first place so you decide to write a program that, for given values of h and k, will determine an optimum set of stamps and the value of n(h,k).

 

Input

Input will consist of several lines, each containing a value for h and k. The file will be terminated by two zeroes (0 0). For technical reasons the sum of h and k is limited to 9. (The President lost his little finger in a shooting accident and cannot count past 9).

 

Output

Output will consist of a line for each value of h and k consisting of the k stamp values in ascending order right justified in fields 3 characters wide, followed by a space and an arrow (->) and the value of n(h,k) right justified in a field 3 characters wide.

 

Sample input

 

3 2
0 0

 

Sample output

 

  1  3 ->  7

 

题意：不好理解啊。。 还是看了别人题解上的题意才懂的。。给定h和k，h代表拥有的邮票张数，k代表有几种面值的邮票。。现在你要确定k个面值。来确保能组合成的连续邮票价值可以到最大。。不是很好理解

举下样例3 和 2.当邮票面值为1 3时。可以组成1 2 3 4 5 6 7 9.连续价值最大到7.所以输出7。假如取1 4.那么组成的为

1 2 3 4 5 6 8 9 12 连续最大到6.要输出较大，所以输出是 1 3-> 7。

思路：从题目中了看出必然有一个面值为1，不然1就组不成了。然后就可以以这个1作为起点。进行2层深搜。一层搜索当前价值能不能被组合成。。一层搜索当价值不能被组合成就添加一种面值。添加的面值区间为[当前能组成的最小面值+1,当前能组成的最大面值+1]。把这几种情况进行搜索。。知道面值数为k结束。

 

#include <stdio.h>
#include <string.h>

int h, k;
int m[10];
int out[10];
int mnum;
int jud;
int maxx;

void judge(int num, int mian, int money)
{
    if (jud == 1)
	return;
    if (mian == money)
    {	
	jud = 1;
	return;
    }
    if (num == h)
	return;
    for (int i = 0; i < mnum; i ++)
    {
	judge(num + 1, mian + m[i], money);
    }
}

void f(int num, int money)
{
    if (num == k)
    {
	if (maxx < money)
	{   
	    maxx = money;
	    for (int i = 0; i < k; i ++)
	    {
		out[i] = m[i];
	    }
	}
	return;
    }
    jud = 0;
    judge(0, 0, money);
    if (jud)
    {
	f(num, money + 1);
    }
    else
    {
	for (int i = 2; i <= money; i ++)
	{
	    m[mnum] = i;
	    mnum ++;
	    f(num + 1, money);
	    mnum --;
	}
    }
}
int main()
{
    while (scanf("%d%d", &h, &k) != EOF && h && k)
    {
	memset(out, 0, sizeof(out));
	memset(m, 0, sizeof(m));
	maxx = 0;
	m[0] = 1;
	mnum = 1;
	f(0, 1);
	for (int i = 0; i < k; i ++)
	    printf("%3d", out[i]);
	printf(" ->%3d\n", maxx - 1);
    }
    return 0;
}