由t<a1<t+1而,结合数列的递推公式可知,当an≥t有an+1=an-t,得a2=a1-t,从而有0<a1-t<1<2<t,即a2<t,同理可得t+1<a3<t+2,1<a4<2,从而有a5=t+2-t-2+a1=a1,可求数列的周期即k的最小值
【解析】
由t<a1<t+1,而当an≥t有an+1=an-t,得a2=a1-t,
又由t<a1<t+1得0<a1-t<1<2<t,即a2<t,
则a3=t+2-a2=t+2-a1+t=2t+2-a1,
又由0<a2<1得t+1<t+2-a2<t+2,即t+1<a3<t+2,
则a4=a3-t=2t+2-a1-t=t+2-a1
又由t+1<a3<t+2得1<a3-t<2,即1<a4<2
则a5=t+2-t-2+a1=a1故最小正周期T=4.
故选B