bzoj2326

首先不难得出递推式f[i]=(f[i-1]*10^k+i) mod m;
f[i]表示接到第i个数时的余数，k表示i的位数
不难想到先按位数穷举位数，然后对于确定的位数，构造矩阵解决
易得出：
f[i]    10^k  1   1      f[i-1] 
 i   =   0    1   1   *    i-1
 1       0    0   1         1
矩阵乘法优化的特点就是有一维特别的大，且这一阶段的值只和上一阶段有关

 var a,b,w,c:array[..,..] of int64;
     f,p:array[..] of int64;
     d:array[..] of longint;
     e:array[..] of int64;
     j,i,l,m:longint;
     x,n:int64;

 procedure mul1;
   var i,j,k:longint;
   begin
     for i:= to  do
       for j:= to  do
       begin
         c[i,j]:=;
         for k:= to  do
           c[i,j]:=(c[i,j]+a[i,k]*b[k,j] mod m) mod m;
       end;
   end;

 procedure mul2;
   var i,k:longint;
   begin
     for i:= to  do
     begin
       f[i]:=;
       for k:= to  do
         f[i]:=(f[i]+c[i,k]*p[k] mod m) mod m;
     end;
   end;

 procedure quick(x:int64);
   var i,j:longint;
   begin
     j:=;
     while x> do
     begin
       inc(j);
       d[j]:=x mod ;
       x:=x div ;
     end;
     fillchar(c,sizeof(c),);
     for i:= to  do
       c[i,i]:=;
     for i:=j downto  do
     begin
       a:=c;
       b:=c;
       mul1;
       if d[i]= then
       begin
         a:=c;
         b:=w;
         mul1;
       end;
     end;
   end;

 begin
   readln(n,m);
   e[]:=;
   l:=;
   for i:= to  do
   begin
     e[i]:=e[i-]*;
     if e[i]>n then
     begin
       l:=i;
       break;
     end;
   end;
   e[]:=e[] mod m* mod m;
   x:=;
   w[,]:=e[] mod m;  w[,]:=;  w[,]:=;
   w[,]:=;  w[,]:=;  w[,]:=;
   w[,]:=;  w[,]:=;  w[,]:=;
   f[]:=;  f[]:=;  f[]:=;
   if l<> then x:=
   else x:=n-;
   p:=f;
   quick(x);
   mul2;
   x:=;
   for i:= to l do
   begin
     p:=f;
     if i<>l then x:=x*
     else x:=n-e[i-]+;  //防止爆int64的
     w[,]:=e[i] mod m;
     quick(x);
     mul2;
   end;
   writeln(f[]);
 end.