【HDU】1709 - The Balance（母函数）

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The Balance

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7474 Accepted Submission(s): 3101

Problem Description

Now you are asked to measure a dose of medicine with a balance and a number of weights. Certainly it is not always achievable. So you should find out the qualities which cannot be measured from the range [1,S]. S is the total quality of all the weights.

Input

The input consists of multiple test cases, and each case begins with a single positive integer N (1<=N<=100) on a line by itself indicating the number of weights you have. Followed by N integers Ai (1<=i<=N), indicating the quality of each weight where 1<=Ai<=100.

Output

For each input set, you should first print a line specifying the number of qualities which cannot be measured. Then print another line which consists all the irrealizable qualities if the number is not zero.

Sample Input

   3
1 2 4
3
9 2 1

Sample Output

   0
2
4 5

Source

HDU 2007-Spring Programming Contest

同样是天平用母函数的问题，思路同上一篇博客。

代码如下：

#include <cstdio>
#include <cstring>
#define CLR(a) memset(a,0,sizeof(a))
int abs(int x)
{
	return x < 0 ? -x : x;
}
int main()
{
	int n;
	int sum;
	int w[111];
	int c[10011];
	int t[10011];
	int ant;
	int num[10011];
	while (~scanf ("%d",&n))
	{
		sum = 0;
		ant = 0;
		CLR(c);
		CLR(t);
		for (int i = 1 ; i <= n ; i++)
		{
			scanf ("%d",&w[i]);
			sum += w[i];
		}
		c[0] = c[w[1]] = 1;
		for (int i = 2 ; i <= n ; i++)
		{
			for (int j = 0 ; j <= sum ; j++)
				for (int k = 0 ; k <= w[i] && k+j <= sum ; k += w[i])
				{
					t[j+k] += c[j];
					t[abs(j-k)] += c[j];
				}
			for (int j = 0 ; j <= sum ; j++)
			{
				c[j] = t[j];
				t[j] = 0;
			}
		}
		for (int i = 1 ; i <= sum ; i++)
		{
			if (c[i] == 0)
				num[ant++] = i;
		}
		if (ant)
		{
			printf ("%d\n",ant);
			for (int i = 0 ; i < ant-1 ; i++)
				printf ("%d ",num[i]);
			printf ("%d\n",num[ant-1]);
		}
		else
			printf ("0\n");
	}
	return 0;
}