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 }

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