题目大意是将给出的n个数分成两堆，使两堆的差尽可能下，且两堆的数的个数不能相差超过1。
一开始是想用f[i,j]表示前i个数中取j个与sum/2的差值最小的值来DP，但发现这是有后效性的，无奈之下只好用了做n次0/1背包的方法来得到在n个数中取n/2个的可能取值，与sum/2最接近的可能值即为所求。
3025033 2008-08-06 18:29:30 Accepted 1880 FPC 00:02.57 2652K IwfWcf@LZOI
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