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[1..3,1..3] of int64;
    f,p:array[1..3] of int64;
    d:array[0..100] of longint;
    e:array[0..20] of int64;
    j,i,l,m:longint;
    x,n:int64;

procedure mul1;
  var i,j,k:longint;
  begin
    for i:=1 to 3 do
      for j:=1 to 3 do
      begin
        c[i,j]:=0;
        for k:=1 to 3 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:=1 to 3 do
    begin
      f[i]:=0;
      for k:=1 to 3 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:=0;
    while x>0 do
    begin
      inc(j);
      d[j]:=x mod 2;
      x:=x div 2;
    end;
    fillchar(c,sizeof(c),0);
    for i:=1 to 3 do
      c[i,i]:=1;
    for i:=j downto 1 do
    begin
      a:=c;
      b:=c;
      mul1;
      if d[i]=1 then
      begin
        a:=c;
        b:=w;
        mul1;
      end;
    end;
  end;

begin
  readln(n,m);
  e[0]:=1;
  l:=19;
  for i:=1 to 18 do
  begin
    e[i]:=e[i-1]*10;
    if e[i]>n then
    begin
      l:=i;
      break;
    end;
  end;
  e[19]:=e[18] mod m*10 mod m;
  x:=0;
  w[1,1]:=e[1] mod m;  w[1,2]:=1;  w[1,3]:=1;
  w[2,1]:=0;  w[2,2]:=1;  w[2,3]:=1;
  w[3,1]:=0;  w[3,2]:=0;  w[3,3]:=1;
  f[1]:=1;  f[2]:=1;  f[3]:=1;
  if l<>1 then x:=8
  else x:=n-1;
  p:=f;
  quick(x);
  mul2;
  x:=9;
  for i:=2 to l do
  begin
    p:=f;
    if i<>l then x:=x*10
    else x:=n-e[i-1]+1;  //防止爆int64的
    w[1,1]:=e[i] mod m;
    quick(x);
    mul2;
  end;
  writeln(f[1]);
end.