f(x+3)≥f(x)+3
f(x+2)≤f(x)+2
f(1)=2
f(x+2)≥f(x-1)+3
又f(x+2)≤f(x)+2
所以f(x-1)+3≤f(x)+2
所以f(x) ≥f(x-1)+1………………….1
又f(x+2)≤f(x)+2
f(x)≤f(x-2)+2
所以f(x-1)+1≤f(x-2)+2
f(x)+1≤f(x-1)+2
f(x)≤f(x-1)+1…………………………….2
由1,2两式可得
f(x)=f(x-1)+1
a2008=f(2008)=f(1)+2007=2009