HDUOJ ----1709

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4737    Accepted Submission(s): 1895

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

 

 

http://acm.hdu.edu.cn/showproblem.php?pid=1709

母函数，有左右算法....很好的题目..

思路...先选着一个点作为‘0’点，然后对左右放置之后的数组（即幂）进行右移动.....最后统计.....很吊的一道母函数呀！！，题目非常新颖.....

 #include<iostream>
 #include<vector>
 #include<cstring>
 #define maxn 20003
 #define gong 10000
 using namespace std;
 int c1[maxn],c2[maxn];
 int save[maxn];
 int main()
 {
     int n,i,sum,j,k,count;
     while(cin>>n)
     {
        vector<int>arr(n);
        memset(c1,,sizeof(c1));
        memset(c2,,sizeof c2);
        memset(save,,sizeof save);
        count=sum=;
        for(i=;i<n;i++)
        {
            cin>>arr[i];
            sum+=arr[i];
        }
        for(i=-arr[];i<=arr[];i+=arr[])
        {
            c1[i+gong]=;
        }
      for(i=;i<n;i++)
      {
          for(j=-sum;j<=sum;j++)
          {
            for(k=-arr[i];k<=arr[i]&&k+j<=sum;k+=arr[i])
            {
              c2[k+j+gong]+=c1[j+gong];
            }
          }

          for(j=-sum+gong;j<=sum+gong;j++)
          {
              c1[j]=c2[j];
              c2[j]=;
          }
      }
        int ans=;
        for(j=;j<=sum;j++)
        {
            if(c1[j+gong]==||c1[gong-j]==)
            {
                save[ans++]=j;
            }
        }
        cout<<ans<<endl;
        if(ans)
        {
          for(i=;i<ans;i++)
          {
              if(i==)
                  cout<<save[i];
              else
                  cout<<" "<<save[i];
          }
           cout<<endl;
        }
     }
     return ;
 }