令 (N=10^{18}-1,sumlimits_{i=0}^{N}f(i)equiv ppmod{a})。
[sumlimits_{i=1}^{N+1}f(i)equiv p+1pmod{a}\
sumlimits_{i=2}^{N+2}f(i)equiv p+2pmod{a}\
vdots\
sumlimits_{i=a-p}^{N+a-p}f(i)equiv p+a-pequiv 0pmod{a}
]
所以答案的 (l=a-p,r=N+a-p)，现在的关键就是求出 (p)。
[egin{aligned}
p&=sumlimits_{i=0}^N f(i)\
&=45 imes 10^{17}+10 imes sumlimits_{i=0}^{10^{17}-1}f(i)\
&=45 imes 10^{17} +45 imes10^{17}+100 imes sumlimits_{i=0}^{10^{16}-1}f(i)\
&=45 imes10^{17} imes 18\
&=81 imes 10^{18}
end{aligned}
]