hduoj1709!【母函数】

/*The Balance
Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5541    Accepted Submission(s): 2243


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 
*/
#include<stdio.h>
#include<string.h>
int main()
{
	int a[110], b[11000], i, j, k, n, sum, c[11000], lc;
	while(scanf("%d", &n) != EOF)
	{
		sum = 0;
		memset(b, 0 , sizeof(b));
		for(i = 0; i < n; i++)
		{
			scanf("%d", &a[i]);
			sum += a[i];
		}
		b[0] = 1;//初始化
		b[a[0]] = 1;//初始化
		for(i = 2; i <= n; i++)
		{
			memset(c, 0, sizeof(c));
			for(j = 0; j <= sum; j++)
			for(k = 0; k + j <= sum && k <= a[i-1]; k += a[i-1])//每个重量的砝码只有一个
			{
				if( k > j) c[k-j] += b[j];//双向称砝码
				else 	 c[j-k] += b[j];
				c[j+k] += b[j];
			}
			for(k  =0; k <= sum ; k++)
			b[k] = c[k];
		}
		int num = 0;
		for(i = 0;i <= sum; i++)
		if( !b[i])
		num++;
		printf("%d\n", num);
		if(num == 0)
		continue;
		else
		{
			j = 0;
			for(i = 0; i <= sum; i++)
			if(!b[i])
			{
				j++;
				if(j != num)
				printf("%d ", i);
				else
				printf("%d\n", i);
			}
		}
	}
	return 0;
}

题意：给你n个砝码，用天秤秤出所能得到的各种质量，输出1到砝码总和之间所不能得到的数字。

注意：砝码可以左右分开放，得到不同的质量数。

这里用的依旧是母函数，难点在于怎么处理左右放砝码，左右放砝码就好比是减法得出一个新的数字。