链接：Link 耗时：0.139s

前言

这道题的主要思路就是打表，看看Fibonacci数列模n几个一循环。但由于这题给的数太大了，从而在细节上耗了很久。在此记录一下：

var    x: qword;    y: longint;begin    x := 1<<64-1;    y := 100;    x := x mod y; //报错201    x := x mod qword(y); //正确end.

Code

var    a,b: qword;    _, n, i, k, cnt: longint;    f: array [1..1000000] of longint;function superMod(a, b: qword; m: longint): longint;var    x: qword;begin    if b = 0 then        exit(1);    x := superMod(a, b shr 1, m);    superMod := x * x mod m;    if odd(b) then        superMod := superMod * a mod m;end;begin    assign(input, 'main.in'); reset(input);    assign(output, 'main.out'); rewrite(output);    readln(_);    while _ > 0 do    begin        dec(_);        readln(a, b, n);        if a = 0 then        begin            writeln(0);            continue;        end;        if n = 1 then        begin            writeln(0);            continue;        end;        f[1] := 1;        f[2] := 1;        cnt := 2;        while not ((f[cnt-1] = 1) and (f[cnt] = 0)) do        begin            inc(cnt);            f[cnt] := (f[cnt-1] + f[cnt-2]) mod n;        end;        //while x > int64(1 <<60) do        //    x := x - int64((cnt << 59));        a := a mod qword(cnt);        k := superMod(a, b, cnt);        writeln(f[k]);    end;    close(output); close(input);end.