2014 AMC 12A Problem 24

If $f_{n-1}(x)=\pm k,$ then $f_n(x)=k-1.$ So if $f_0(x)=\pm k$ for a nonnegative integer $k,$ then $f_k(x)=0,$ after which the sequence alternates $0,-1,0,\ldots$ Thus $f_{100}(x)=0$ exactly when $f_0(x)=2k$ for some integer $-50\le k\le50.$ Now $f_0(x)=x+|x-100|$ $-|x+100|$ equals $x+200$ for $x\lt-100,$ $-x$ for $-100\le x\lt100,$ and $x-200$ for $x\ge100.$ Its graph is piecewise linear with turning points $(-100,100)$ and $(100,-100).$ A line $y=2k$ meets this graph three times for $-49\le k\le49$ and twice for $k=\pm50.$ The total is $99\cdot3+2\cdot2=301.$ Thus, the correct answer is C.