整数的平均值

对于两个整数x,y,如果用 (x+y)/2 求平均值,会产生溢出,因为 x+y 可能会大于INT_MAX,但是我们知道它们的平均值是肯定不会溢出的,我们用如下算法:

#define AVE(x,y) ((x)&(y))+(((x)^(y))>>1)

 

http://blog.csdn.net/oanqoanq/article/category/938009

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