51问答网 > 有一个运算程序，

有一个运算程序，

2025-03-15 07:31:51
推荐回答（1个）
回答1：

2010⊕2009=（ -2005 ）

1⊕1=2
2⊕1 = 2+1 = 3
2⊕2 = 3 - 2 = 1

也就是⊕前后各增加1时，值减少1。
因此
2009⊕2009 = 2 - (2009-1)*1 = 2 - 2008 = -2006
2010⊕2009 = -2006 + 1 = -2005

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