bzoj 1876 高精

　　首先我们知道，对于两个数a,b，他们的gcd情况有如下形式的讨论

　　当a为奇数，b为偶数的时候gcd(a,b)=gcd(a div 2,b)

　　当b为奇数，a为偶数的时候gcd(a,b)=gcd(a,b div 2)

　　当a为偶数，b为偶数的时候gcd(a,b)=2*gcd(a div 2,b div 2)

　　当a为奇数，b为奇数的时候，根据欧几里德定律，有gcd(a,b)=gcd(a-b,b) (a>b)时

　　那么这道题就变成了不断地缩小a,b的范围了。直接高精就行了。当然数据为1,10^1000的时候会tle，题目比较良心没有这样的数据。

高精写渣了。

/**************************************************************
    Problem:
    User: BLADEVIL
    Language: Pascal
    Result: Time_Limit_Exceed
****************************************************************/

//By BLADEVIL
var
    s1, s2                      :ansistring;
    f1, f2                      :boolean;
    ans                         :ansistring;
    a, b, c                     :array[..] of longint;
    i                           :longint;
    doit                        :longint;

function max(s1,s2:ansistring):boolean;
begin
    if length(s1)>length(s2) then exit(true);
    if (length(s1)=length(s2)) and (s1>s2) then exit(true);
    exit(false);
end;

function divid(s:ansistring):ansistring;
var
    i                           :longint;
    len                         :longint;
    ss                          :ansistring;
begin
    fillchar(a,sizeof(a),);
    len:=length(s);
    for i:= to len do a[(len-i) div +]:=a[(len-i) div +]*+ord(s[i])-;
    len:=(len+) div ;
    for i:=len downto  do
        if a[i] mod = then
            a[i]:=a[i] div  else
            begin
                a[i]:=a[i] div ;
                a[i-]:=a[i-]+;
            end;
    a[]:=a[] div ;
    divid:='';
    for i:=len downto  do
    begin
        str(a[i],ss);
        if a[i]< then divid:=divid+'';
        if a[i]< then divid:=divid+'';
        if a[i]< then divid:=divid+'';
        if a[i]< then divid:=divid+'';
        if a[i]< then divid:=divid+'';
        if a[i]< then divid:=divid+'';
        divid:=divid+ss;
    end;
    while (divid[]='') and (length(divid)>) do delete(divid,,);
end;

function jian(s1,s2:ansistring):ansistring;
var
    len1, len2                  :longint;
    ss                          :ansistring;
    i                           :longint;
begin
    fillchar(a,sizeof(a),);
    fillchar(b,sizeof(b),);
    fillchar(c,sizeof(c),);
    len1:=length(s1);
    for i:= to len1 do a[(len1-i) div +]:=a[(len1-i) div +]*+ord(s1[i])-;
    len2:=length(s2);
    for i:= to len2 do b[(len2-i) div +]:=b[(len2-i) div +]*+ord(s2[i])-;
    len1:=(len1+) div ;
    len2:=(len2+) div ;
    for i:= to len1 do
    begin
        c[i]:=c[i]+a[i]-b[i];
        if c[i]< then
        begin
            c[i]:=c[i]+;
            c[i+]:=c[i+]-;
        end;
    end;
    jian:='';
    for i:=len1 downto  do
    begin
        str(c[i],ss);
        if c[i]< then jian:=jian+'';
        if c[i]< then jian:=jian+'';
        if c[i]< then jian:=jian+'';
        if c[i]< then jian:=jian+'';
        if c[i]< then jian:=jian+'';
        if c[i]< then jian:=jian+'';
        jian:=jian+ss;
    end;
    while (jian[]='') and (length(jian)>) do delete(jian,,);
end;

function mul(s:ansistring):ansistring;
var
    len                         :longint;
    i                           :longint;
    ss                          :ansistring;
begin
    len:=length(s);
    fillchar(a,sizeof(a),);
    for i:= to len do a[(len-i) div +]:=a[(len-i) div +]*+ord(s[i])-;
    len:=(len+) div ;
    for i:= to len do a[i]:=a[i]*;
    for i:= to len do
    begin
        a[i+]:=a[i+]+a[i] div ;
        a[i]:=a[i] mod ;
    end;
    inc(len);
    mul:='';
    for i:=len downto  do
    begin
        str(a[i],ss);
        if a[i]< then mul:=mul+'';
        if a[i]< then mul:=mul+'';
        if a[i]< then mul:=mul+'';
        if a[i]< then mul:=mul+'';
        if a[i]< then mul:=mul+'';
        if a[i]< then mul:=mul+'';
        mul:=mul+ss;
    end;
    while (mul[]='') and (length(mul)>) do delete(mul,,);
end;

begin
    readln(s1);
    while (s1[]='') and (length(s1)>) do delete(s1,,);
    readln(s2);
    while (s2[]='') and (length(s2)>) do delete(s2,,);
    doit:=;
    while s1<>s2 do
    begin
        if ord(s1[length(s1)]) mod = then f1:=true else f1:=false;
        if ord(s2[length(s2)]) mod = then f2:=true else f2:=false;
        if f1 and f2 then
        begin
            s1:=divid(s1);
            s2:=divid(s2);
            inc(doit);
        end else
        begin
            if f1 then s1:=divid(s1);
            if f2 then s2:=divid(s2);
            if (not f1) and (not f2) then
                if max(s1,s2) then s1:=jian(s1,s2) else s2:=jian(s2,s1);
        end;
    end;
    ans:=s1;
    for i:= to doit do ans:=mul(ans);
    writeln(ans);
end.