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(c1,0,sizeof(c1));
       memset(c2,0,sizeof c2);
       memset(save,0,sizeof save);
       count=sum=0;
       for(i=0;i<n;i++)
       {
           cin>>arr[i];
           sum+=arr[i];
       }
       for(i=-arr[0];i<=arr[0];i+=arr[0])
       {
           c1[i+gong]=1;
       }
     for(i=1;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]=0;
         }
     }
       int ans=0;
       for(j=0;j<=sum;j++)
       {
           if(c1[j+gong]==0||c1[gong-j]==0)
           {   
               save[ans++]=j;
           }
       }
       cout<<ans<<endl;
       if(ans)
       {
         for(i=0;i<ans;i++)
         {
             if(i==0)
                 cout<<save[i];
             else
                 cout<<" "<<save[i];
         }
          cout<<endl;       
       }
    }
    return 0;
}