Dividing
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3850 Accepted Submission(s): 1035
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.
把若干不同價值的石頭分爲價值相等的兩份 , 可以轉化爲多重揹包問題 , 把物品的價值和花費都設爲石頭的價值 , 如果能夠裝滿一個容量爲其總價值的一半的揹包 , 那麼就可以平均分爲兩份 . 這裏的多重揹包使用的是單調隊列優化 DP 的算法 .
代碼如下 :
