HDOJ 1709 The Balance(母函数)

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4497    Accepted Submission(s): 1808

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

 

 

Source

HDU 2007-Spring Programming Contest

 

 

Recommend

lcy

 

 

 1 //对每个砝码可以不用，也可以放到左边或右边。。。。。
 2 
 3 #include 
 4 #include 
 5 #include 
 6 
 7 using namespace std;
 8 
 9 int n,m;
10 int a[111];
11 int c1[22222],c2[22222];
12 const int cc=10001;
13 
14 int main()
15 {
16 while(scanf("%d",&n)!=EOF)
17 {
18     int m=0;
19     for(int i=1;i<=n;i++)
20     {
21         scanf("%d",&a[i]);
22         m+=a[i];
23     }
24 
25     memset(c1,0,sizeof(c1));
26     memset(c2,0,sizeof(c2));
27 
28     for(int i=-a[1];i<=a[1];i+=a[1])
29     {
30         c1[i+cc]=1;
31     }
32 
33     for(int i=2;i<=n;i++)
34     {
35         for(int j=-m;j<=m;j++)
36         {
37             for(int k=-a[i];k+j<=m&&k<=a[i];k+=a[i])
38             {
39                 c2[k+j+cc]+=c1[j+cc];
40             }
41         }
42         for(int j=-m;j<=m;j++)
43         {
44             c1[j+cc]=c2[j+cc];
45             c2[j+cc]=0;
46         }
47     }
48 
49     int aas[10000]={0};
50     int ans=0;
51     for(int i=0;i<=m;i++)
52     {
53         if(c1[i+cc]||c1[-i+cc]) ;
54         else
55         {
56             aas[ans]=i;
57             ans++;
58         }
59     }
60 
61     printf("%d\n",ans);
62     for(int i=0;i)
63     {
64         if(i==0)
65             printf("%d",aas[i]);
66         else
67             printf(" %d",aas[i]);
68     }
69     if(ans)
70     putchar(10);
71 
72 }
73 
74     return 0;
75 }

 

转载于:https://www.cnblogs.com/CKboss/p/3165229.html