Problem Description
Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value.
Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.
 

Input
Each line in the input describes one collection of marbles to be divided. The lines consist of six non-negative integers n1, n2, ..., n6, where ni is the number of marbles of value i. So, the example from above would be described by the input-line ``1 0 1 2 0 0''. The maximum total number of marbles will be 20000.

The last line of the input file will be ``0 0 0 0 0 0''; do not process this line.
 

Output
For each colletcion, output ``Collection #k:'', where k is the number of the test case, and then either ``Can be divided.'' or ``Can't be divided.''.

Output a blank line after each test case.
 

Sample Input
1 0 1 2 0 0

1 0 0 0 1 1

0 0 0 0 0 0
 

Sample Output
Collection #1:

Can't be divided.

Collection #2:

Can be divided.

好難，我自閉了。

題意：玻璃球價值從1-6數量給你，問你能不能均分成等價的兩份。

思路：多重揹包二進位制優化模板。揹包容量是總價的一半，看看能不能裝滿。資料很大，要用二進位制優化轉01揹包。下面程式碼詳解。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 
 6 int dp[120005],a[10];
 7 int m;
 8 
 9 void completepack(int c,int w){//正向 
10     for(int i=c;i<=m;i++){
11         if(dp[i-c]+w>dp[i]) dp[i]=dp[i-c]+w;
12     }
13 }
14 void zeroonepack(int c,int w){//逆向 
15     for(int i=m;i>=c;i--){
16         if(dp[i-c]+w>dp[i]) dp[i]=dp[i-c]+w;
17     }
18 }
19 //這樣1,2,4,8的塞肯定能覆蓋所有塞的數量。 
20 void multiplepack(int c,int w,int t){//c價值、w重量、t該種數量 
21     if(t*w>=m) completepack(c,w);//如果t*w該種總重大於容量 
22     else{//類似於快速冪的操作
23         int k=1;
24         while(k<t){//如果小於該種剩餘
25             zeroonepack(c*k,w*k);//假設新物品（k（1,2,4,8...）個該種物品拼成） 
26             t-=k;//更新該種剩餘
27             k<<=1;//k*=2
28         }
29         zeroonepack(c*t,w*t);//剩餘的合成一個新物品 
30     }
31 }
32 
33 int main(){
34     int cnt=0;
35     while(1){
36         m=0;
37         for(int i=1;i<=6;i++){
38             scanf("%d",&a[i]);
39             m+=a[i]*i;
40         }
41         if(m==0) break;
42         printf("Collection #%d:\n",++cnt);
43         
44         if(m&1){//m是奇數肯定不能均分 
45         printf("Can't be divided.\n\n");
46         continue;
47         }
48         m>>=1;//m取總量的一半 
49         memset(dp,0,sizeof(dp));
50         for(int i=1;i<=6;i++)
51             multiplepack(i,i,a[i]);//重量和價值一樣 
52         
53         if(dp[m]==m) printf("Can be divided.\n\n");
54         else printf("Can't be divided.\n\n");    
55                 
56     }
57     return 0;
58 }