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.....
1 #include<iostream>
2 #include<vector>
3 #include<cstring>
4 #define maxn 20003
5 #define gong 10000
6 using namespace std;
7 int c1[maxn],c2[maxn];
8 int save[maxn];
9 int main()
10 {
11 int n,i,sum,j,k,count;
12 while(cin>>n)
13 {
14 vector<int>arr(n);
15 memset(c1,0,sizeof(c1));
16 memset(c2,0,sizeof c2);
17 memset(save,0,sizeof save);
18 count=sum=0;
19 for(i=0;i<n;i++)
20 {
21 cin>>arr[i];
22 sum+=arr[i];
23 }
24 for(i=-arr[0];i<=arr[0];i+=arr[0])
25 {
26 c1[i+gong]=1;
27 }
28 for(i=1;i<n;i++)
29 {
30 for(j=-sum;j<=sum;j++)
31 {
32 for(k=-arr[i];k<=arr[i]&&k+j<=sum;k+=arr[i])
33 {
34 c2[k+j+gong]+=c1[j+gong];
35 }
36 }
37
38 for(j=-sum+gong;j<=sum+gong;j++)
39 {
40 c1[j]=c2[j];
41 c2[j]=0;
42 }
43 }
44 int ans=0;
45 for(j=0;j<=sum;j++)
46 {
47 if(c1[j+gong]==0||c1[gong-j]==0)
48 {
49 save[ans++]=j;
50 }
51 }
52 cout<<ans<<endl;
53 if(ans)
54 {
55 for(i=0;i<ans;i++)
56 {
57 if(i==0)
58 cout<<save[i];
59 else
60 cout<<" "<<save[i];
61 }
62 cout<<endl;
63 }
64 }
65 return 0;
66 }