Dividing
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20691 Accepted Submission(s): 5827

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 eitherCan 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.

題解：

背包問題，二進制優化，對於價值為val有num個，可以將其分成1,2,4,8......這樣的二進制價值，可以保證如果可以均分，是一定

可以的，如果可以均分，那麽保證可以分出，如11 可以分出1,2,4,6，不可以均分，12分出1,2,4,5拿1,5或者2,4就均分了。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<cstdlib>
 7 using namespace std;
 8 
 9 int Case=0;
10 int cont=0;
11 int a[10],val[107];
12 bool f[60007];
13 
14 int main()
15 {
16     while(~scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])&&(a[1]+a[2]+a[3]+a[4]+a[5]+a[6]))
17     {
18         printf("Collection #%d:\n",++Case);
19         int sum=0,mid=0;
20         for (int i=1;i<=6;i++)
21             sum=sum+a[i]*i;
22         if (sum%2)
23         {
24             printf("Can‘t be divided.\n\n");
25             continue;
26         }
27         mid=sum/2;
28         memset(f,0,sizeof(f));
29         f[0]=1;cont=0;
30         for (int i=1;i<=6;i++)
31         {
32             int t=i;a[i]=a[i]*i;
33             while(a[i]>=t)
34             {
35                 val[++cont]=t;
36                 a[i]-=t;
37                 t*=2;
38             }
39             if (a[i]>0) val[++cont]=a[i];
40         }
41         f[0]=1;
42         for (int i=1;i<=cont;i++)
43             for (int j=mid;j>=val[i];j--)
44                 if (f[j-val[i]]) f[j]=1;
45         if (f[mid]) printf("Can be divided.\n\n");
46         else printf("Can‘t be divided.\n\n");    
47     }
48 }

hdu1059（背包dp二進制優化）