HDOJ_1709 The Balance 解题报告

 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

 

 

解题思路：用母函数去求解，只是这个增加了可能有减得情况，主要还是组合的一个思想。

代码如下：

Code:

1. #include<iostream>  
2. using namespace std;  
3. const int Max1(10001);  
4. const int Max2(101);  
5. int num[Max1];  
6. int temp[Max1];  
7. int data[Max2];  
8. int main()  
9. {  
10.     int n;  
11.     while(cin>>n)  
12.     {  
13.         int sum=0;  
14.         num[0]=1;  
15.         for(int i=1;i<=n;i++)  
16.         {  
17.             cin>>data[i];  
18.             sum+=data[i];  
19.               
20.         }  
21.         for(i=0;i<=sum;++i)  
22.         {  
23.             num[i]=0;  
24.             temp[i]=0;  
25.         }  
26.         num[0]=1;  
27.       //用到两重循环，从来到尾的找即可  
28.         for(i=1;i<=n;i++)  
29.         {  
30.             for(int j=0;j+data[i]<=sum;j++)           
31.             {  
32.                 if(num[j]==1)  
33.                 {  
34.                     temp[j]=1;  
35.                     temp[j+data[i]]=1;  
36.                     temp[abs(j-data[i])]=1;    
37.                 }  
38.             }  
39.             for(j=0;j<=sum;j++)  
40.             {  
41.                 num[j]=temp[j];  
42.                 temp[j]=0;  
43.             }  
44.         }  
45.         int step=0;  
46.         int j;  
47.         for(i=1,j=0;i<=sum;i++)  
48.         {  
49.             if(num[i]==0)  
50.             {  
51.                 step++;  
52.                 temp[j]=i;  
53.                 j++;  
54.             }  
55.         }  
56.         cout<<step<<endl;  
57.         if(step!=0)  
58.         {  
59.             for(i=0;i<j;i++)  
60.             {  
61.                 if(i!=0)  
62.                     cout<<' ';  
63.                 cout<<temp[i];  
64.             }  
65.             cout<<endl;  
66.         }  
67.     }  
68.     return 0;  
69. }