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

解题思路：

用天平称重时，可以有两种方法。一种是砝码和物体分别放在两侧；

另一种是砝码和物体放在同一侧，另一侧再放砝码，此时和物体在同一侧的砝码相当于是负重量，这是本题的关键

首先，砝码重量可能相等，要统计各自的数量

然后再套用母函数

最后，统计不能称量的

程序代码：

#include<cstdio>

#include<cmath>

#include<cstring>

#include<algorithm>

#define W 10005

using namespace std;

int b[W],c[W];

int main()

{

 int a[],d[];

 int N,i,t,n,j,len,r,p,k;

 while(scanf("%d",&N)!=EOF)

 {

  for(i=;i<N;i++)

  scanf("%d",&a[i]);

  sort(a,a+N);

  d[]=t=a[];n=;j=;

  for(i=;i<N;i++)

  {

   if(a[i]==t){n++;}

   else {d[j]=n;n=;a[j++]=t;t=a[i];}

  }

  a[j]=t;

  d[j]=n;

  memset(b,,sizeof(b));

  memset(c,,sizeof(c));

  for(i=;i<=d[]*a[];i+=a[])

   c[i]=;len=d[]*a[];

  for(i=;i<=j;i++)

  {

   for(p=;p<=len;p++)

   for(k=;k<=d[i]*a[i];k+=a[i])

   {

    b[p+k]+=c[p];

       if(k!=&&p!=)

        {r=fabs(p-k);

    b[r]+=c[p];}

   }

   len+=d[i]*a[i];

   for(k=;k<=len;k++)

   {c[k]=b[k];b[k]=;}

  }

  n=;

  for(i=,k=;i<=len;i++)

  if(c[i]==) b[k++]=i;

  if(k==)printf("0\n");

  else {

   printf("%d\n",k);

  for(i=;i<k;i++)

  { if(i!=)printf(" ");

   printf("%d",b[i]);

  }

   printf("\n");

  }

 }

 return ;

}