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
1 var a,b,w,c:array[1..3,1..3] of int64; 2 f,p:array[1..3] of int64; 3 d:array[0..100] of longint; 4 e:array[0..20] of int64; 5 j,i,l,m:longint; 6 x,n:int64; 7 8 procedure mul1; 9 var i,j,k:longint; 10 begin 11 for i:=1 to 3 do 12 for j:=1 to 3 do 13 begin 14 c[i,j]:=0; 15 for k:=1 to 3 do 16 c[i,j]:=(c[i,j]+a[i,k]*b[k,j] mod m) mod m; 17 end; 18 end; 19 20 procedure mul2; 21 var i,k:longint; 22 begin 23 for i:=1 to 3 do 24 begin 25 f[i]:=0; 26 for k:=1 to 3 do 27 f[i]:=(f[i]+c[i,k]*p[k] mod m) mod m; 28 end; 29 end; 30 31 32 procedure quick(x:int64); 33 var i,j:longint; 34 begin 35 j:=0; 36 while x>0 do 37 begin 38 inc(j); 39 d[j]:=x mod 2; 40 x:=x div 2; 41 end; 42 fillchar(c,sizeof(c),0); 43 for i:=1 to 3 do 44 c[i,i]:=1; 45 for i:=j downto 1 do 46 begin 47 a:=c; 48 b:=c; 49 mul1; 50 if d[i]=1 then 51 begin 52 a:=c; 53 b:=w; 54 mul1; 55 end; 56 end; 57 end; 58 59 begin 60 readln(n,m); 61 e[0]:=1; 62 l:=19; 63 for i:=1 to 18 do 64 begin 65 e[i]:=e[i-1]*10; 66 if e[i]>n then 67 begin 68 l:=i; 69 break; 70 end; 71 end; 72 e[19]:=e[18] mod m*10 mod m; 73 x:=0; 74 w[1,1]:=e[1] mod m; w[1,2]:=1; w[1,3]:=1; 75 w[2,1]:=0; w[2,2]:=1; w[2,3]:=1; 76 w[3,1]:=0; w[3,2]:=0; w[3,3]:=1; 77 f[1]:=1; f[2]:=1; f[3]:=1; 78 if l<>1 then x:=8 79 else x:=n-1; 80 p:=f; 81 quick(x); 82 mul2; 83 x:=9; 84 for i:=2 to l do 85 begin 86 p:=f; 87 if i<>l then x:=x*10 88 else x:=n-e[i-1]+1; //防止爆int64的 89 w[1,1]:=e[i] mod m; 90 quick(x); 91 mul2; 92 end; 93 writeln(f[1]); 94 end.
