HDU 1709

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3218    Accepted Submission(s): 1280

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

 

这题其实思路是非常清晰的，就是应该要分成两个步骤来求解，（1）正常的求母函数的过程与方法（2）由于两两和值的差值也是能够被称出的，因此我们必须要把这一部分也求出来。

就我个人，（1）自然没什么可讲，但一开始（2）步骤想的太过繁琐，结果。。。然后在他人启发下，我们可以设置一个数组，将所有的差值的全存入其中，然后用遍历的思想去找出在此之外的。（其实我的想法也是这样，只不过，更杂乱一些，然后遗漏了一些本应有的）。

其实，母函数是很模板化的。。。我感觉

 

#include"iostream"
using namespace std;
int main()
{
    int c1[10005],c2[10005],a[10005],b[10005],c[10005];
    int i,j,k,n,sum,p;
    while(cin>>n)
    {
        sum=0;
        for(i=0;i<n;i++)
        {
            cin>>a[i];
            sum+=a[i];
        }
        for(i=1;i<=sum;i++)
        {
            c1[i]=0;
            c2[i]=0;
            b[i]=0;
            c[i]=0;
        }
        for(i=0;i<=1;i++)
            c1[i*a[0]]=a[0];
        for(i=1;i<n;i++)
        {
            for(j=0;j<=sum;j++)
                for(k=0;k*a[i]+j<=sum&&k<=1;k++)
                    c2[k*a[i]+j]+=c1[j];
            for(j=0;j<=sum;j++)
            {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }       //到这里完成第一部分
        for(i=sum;i>0;i--)
        {
            if(!c1[i])
                continue;
            for(j=1;j<i;j++)
            {
                if(!c1[j])
                    continue;
                b[i-j]=1;
            }
        }//这里是把能有的差值用这种方式找出来
        p=0;
        for(i=1;i<=sum;i++)
        {
            if(!c1[i]&&!b[i])//s数组中为1的是能够被称出来的（出了data中的以外）
                c[p++]=i;
        }
        cout<<p<<endl;
        if(p)
        {
            for(i=0;i<p-1;i++)
                cout<<c[i]<<' ';
            cout<<c[i]<<endl;
        }
    }
    return 0;
}