先找规律，各个数除以5的余数是1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1,0 ,1,1,2……20个为一个序列，总共100组序列外加两个余数之和是 4000+2所以它们的和除以5的余数是2
a1=a2=1,a(n+2)=a(n+1)+anan=(1/√5)*{[(1+√5)/2]^n - [(1-√5)/2]^n}
余数是 4001
