9-28 解题报告

第一题：fibo矩阵快速幂

无压力，给信心的

type
        matrix=array[1..2,1..2] of qword;
var
        c,cc:matrix;
        t,bili:longint;
        n,p:int64;
        function multiply(x,y:matrix):matrix;
        var
        temp:matrix;
        begin
                temp[1,1]:=(x[1,1]*y[1,1]+x[1,2]*y[2,1]) mod p;
                temp[1,2]:=(x[1,1]*y[1,2]+x[1,2]*y[2,2]) mod p;
                temp[2,1]:=(x[2,1]*y[1,1]+x[2,2]*y[2,1]) mod p;
                temp[2,2]:=(x[2,1]*y[1,2]+x[2,2]*y[2,2]) mod p;
        exit(temp);
        end;

        function getcc(n:int64):matrix;
        var
        temp:matrix;
        t:int64;
        begin
                if n=1 then exit(c);
                t:=n div 2;
                temp:=getcc(t);
                temp:=multiply(temp,temp);
        if odd(n) then exit(multiply(temp,c))
        else exit(temp);
        end;


        procedure init;
        begin
        readln(n,p);
        c[1,1]:=1;
        c[1,2]:=1;
        c[2,1]:=1;
        c[2,2]:=0;
        if n=0 then
        begin
            writeln(0);
            exit;
        end;
        if n=1 then
        begin
                writeln(1 mod p);
                exit;
        end;
        if n=2 then
        begin
                writeln(1 mod p);
                exit;
        end;
        cc:=getcc(n-2);
        end;

        procedure work;
        begin
                writeln((cc[1,1]+cc[1,2]) mod p);
                end;

        begin
                assign(input,'eins.in');
                reset(input);
                assign(output,'eins.out');
                rewrite(output);
                readln(t);
                for bili:=1 to t do
                begin
                init;
                if (n>2) then work;
                end;
                close(input);
                close(output);
        end.

第二题：树状数组
把求和变成求xor，剩下模板，再给信心

var
        a,f:array[0..200010] of longint;
        l,r,x,y,z,m,n,i,j:longint;

        function lowbit(x:longint):longint;
        begin
                exit(x and (-x));
                end;

        procedure build(pos,x:longint);
        var i:longint;
        begin
                i:=pos;
        while i<=n do
        begin
                f[i]:=f[i] xor x;
                inc(i,lowbit(i));
        end;
        end;

        function sum(x:longint):longint;
        var
        tail,s:longint;
        begin
                tail:=x;
                s:=0;
                while tail>0 do
                begin
                        s:=s xor f[tail];
                        dec(tail,lowbit(tail));
                end;
                sum:=s;
        end;

        procedure change(x,y:longint);
        var p:longint;
        begin
        p:=x;
        while p<=n do
        begin
                f[p]:=f[p] xor a[x] xor y;
                inc(p,Lowbit(p));
        end;
        a[x]:=y;
        end;

        begin
        assign(input,'zwei.in');
        reset(input);
        assign(output,'zwei.out');
        rewrite(output);
                readln(n,m);
                for i:=1 to n do
                begin
                        read(a[i]);
                        build(i,a[i]);
                end;

                for i:=1 to m do
                begin
                        read(z,x,y);
                        if z=0 then change(x,y);
                        if z=1 then writeln(sum(y) xor sum(x-1));
                end;
                close(input);
                close(output);
        end.

第三题：
欧拉函数，写了好几次才过，后来发现快速幂写错了（捂脸= =）

var
        i,t,n,mo,ans,k:int64;
        bilibili:longint;



        function gcd(x,y:int64):int64;
        begin
                if y=0 then exit(x);
                exit(gcd(y,x mod y));
        end;

        function phi(k:int64):int64;
        var i,ret:int64;
        begin
                ret:=1;
                i:=2;
                while i*i<=k do
                begin
                        if k mod i=0 then
                        begin
                                ret:=ret*(i-1);
                                k:=k div i;
                                while k mod i=0 do begin ret:=ret*i; k:=k div i; end;
                        end;
                        inc(i);
                end;
                if k>1 then ret:=ret*(k-1);
                exit(ret);
        end;

        function pow(a,b:int64):int64;
        var ret,i:int64;
        begin
                ret:=1;
                while b<>0 do
                begin
                        if b and 1<>0 then begin ret:=(ret*a) mod mo;

                        end;
                        a:=(a*a) mod mo;
                        b:=b>>1;
                end;
                exit(ret mod mo);
        end;

        begin
            assign(input,'drei.in');
            reset(input);
            assign(output,'drei.out');
            rewrite(output);
                readln(t);
                for bilibili:=1 to t do
                begin
                        readln(n,mo);
                        if mo=1 then begin writeln(-1); continue; end;
                        if n=0 then begin writeln(-1); continue; end;
                        if n=1 then begin writeln(1); continue; end;
                        if gcd(n,mo)>1 then begin writeln(-1); continue; end;
                        ans:=phi(mo);
                        k:=ans;
                        i:=2;
                        while i*i<=k do
                        begin
                                if k mod i=0 then
                                begin
                                        if (pow(n,i)=1) and (i<ans) then ans:=i;
                                        if (pow(n,k div i)=1) and (k div i<ans) then ans:=k div i;
                                end;
                                inc(i);
                        end;
                        writeln(ans);
                end;
            close(input);
            close(output);
        end.

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