The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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
<span style="font-family: 'Microsoft YaHei', 微软雅黑, Lucida, Verdana, 'Hiragino Sans GB', STHeiti, 'WenQuanYi Micro Hei', SimSun, sans-serif; font-size: 14px; line-height: 24px;">这里比较特殊的一点是砝码可以放在天枰的左右两端，我们可以在c2[j+k]+=c1[j]</span><br style="font-family: 'Microsoft YaHei', 微软雅黑, Lucida, Verdana, 'Hiragino Sans GB', STHeiti, 'WenQuanYi Micro Hei', SimSun, sans-serif; font-size: 14px; line-height: 24px;" /><span style="font-family: 'Microsoft YaHei', 微软雅黑, Lucida, Verdana, 'Hiragino Sans GB', STHeiti, 'WenQuanYi Micro Hei', SimSun, sans-serif; font-size: 14px; line-height: 24px;">后加多一句c2[abs(j-k)]+=c[j]...即可,假设原来的砝码都放在右端，则可以把新加的砝码放在左端，得到新重量。</span>

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
int main()
{
	int c1[15000],c2[15000],a[150],count[15000];
	int n,i,j,k,s,counts;
	while(~scanf("%d",&n))
	{
		int s=0;
		for(i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			s+=a[i];
		}
		memset(c1,0,sizeof(c1));
		memset(c2,0,sizeof(c2));

		c1[0]=c1[a[0]]=1;
		int end=a[0];
		for(i=2;i<=n;i++)
		{

			for(j=0;j<=end;j++)
			{
				for(k=0;j+k<=s && k<=a[i-1];k+=a[i-1])
				{
					c2[j+k]+=c1[j];
					c2[abs(j-k)]+=c1[j];//系数可能为负数
				}
			}
			end+=a[i-1];
			memcpy(c1,c2,sizeof(c1));
			memset(c2,0,sizeof(c2));

		}
		int cnt=0;
		for(i=0;i<=s;i++)
		{
			if(c1[i]==0)
			{

				c2[cnt++]=i;
			}

		}
		if(cnt==0)
            printf("0\n");
		else
		{
	    	printf("%d\n",cnt);
			for(i=0;i<cnt-1;i++)
			{
				printf("%d ",c2[i]);
			}
			printf("%d\n",c2[i]);
		}
	}
	return 0;
}