题解

74 条题解

  • 0
    @ 2017-06-17 17:01:53

    代码巨丑无比,唉,模拟题就是应该思路清晰,不要为了快,否则出了问题找起来超级麻烦,还不一定有慢慢写写的快,emmmmmm,这次就是血的教训。。。55555555555555555,还有一点就是5-8位和1-4位可以执行同一个函数。。

    #include <iostream>
    #include <cstring>
    using namespace std;
    char num[50];
    char zheng[50];
    void read(char* wan,int flag)
    {
        if(wan[0]!='0')
        {
            cout<<wan[0]<<'Q';
            flag=1;
        }
        if(wan[1]!='0')
        {
            if(wan[0]=='0'&&flag)
            {
                cout<<wan[0];
            }
            cout<<wan[1]<<'B';
            //flag=1;
        }
        if(wan[2]!='0')
        {
            if(wan[1]=='0'&&flag)
            cout<<wan[1];
            cout<<wan[2]<<'S';
        }
        if(wan[3]!='0'&&wan[2]=='0'&&flag)
        {
            cout<<wan[2]<<wan[3];
        }
        if(wan[3]!='0'&&wan[2]=='0'&&!flag)
        {
            cout<<wan[3];
        }
        if(wan[3]!='0'&&wan[2]!='0')
        {
            //if(wan[1]=='0')
            //cout<<wan[1];
            cout<<wan[3];
        }
        return;
    } 
    int main(void)
    {
        int count1=0,count2=0,count3=0;
        int length=0;
        
        char wan[5];
        char wan2[5];
        char wan1[5];
        for(int i=0;i<5;i++)
        {
            wan[i]='0';
            wan1[i]='0';
            wan2[i]='0';
        }
        for(int i=0;i<50;i++)
        {
            zheng[i]='\0';
        }
        cin>>num;
        //sprintf(num,"%llf",m);
        //cout<<num<<endl;
        //cout<<strlen(num)<<endl;
        //cout<<"*****"<<endl;
        length=strlen(num);
        int flag3=0;
        for(int i=0;i<length;i++)
        {
            if(num[i]=='.')
            {
                flag3=1;
                count3=i;
            }
        }
        
        if(num[0]=='-')
        {
            cout<<'F';
            count1++;
        }
        else if(num[0]=='+')
        {
            count1++;
        }
        int flag=0;
        for(int i=count1;i<length;i++)
        {
            if(num[i]=='.')
            {
                break;
            }
            if(num[i]!='0')
            {
                flag=1;
            }
            if(flag)
            {
                zheng[count2]=num[i];
                count2++;
            }
        }
        //cout<<zheng<<"********"<<endl;
        //cout<<"*********"<<endl;
        int ptr=0;
        //cout<<count2<<endl;
        //Y
        if(count2==9)
        {
            cout<<zheng[0]<<'Y';
            ptr++;
        }
        //W
        if(count2>4)
        {
            int index=count2-4-ptr;
            for(int i=0;i<index;i++)
            {
                wan[3-i]=zheng[count2-1-4-i];
            }
            read(wan,ptr);
            if(wan[1]!='0'||wan[0]!='0'||wan[2]!='0'||wan[3]!='0')
            cout<<'W';
            for(int i=0;i<4;i++)
            {
                wan2[3-i]=zheng[count2-1-i];
            }
            read(wan2,1);
        }
        if(count2<=4)
        {
            for(int i=0;i<count2;i++)
            {
                wan1[3-i]=zheng[count2-1-i];
            }
            read(wan1,0);
        }
        if(count2==0)
        {
            cout<<'0';
        }
        //cout<<"**********"<<endl;
        //cout<<length<<endl;
        if(flag3&&!((num[length]=='\0')&&num[length-1]=='.'))
        {
            cout<<'D'; 
            for(int i=count3+1;i<length;i++)
            {
                cout<<num[i];
            }
        }
        
    }
    
  • 0
    @ 2016-09-15 20:35:50

    QaQ这辈子最讨厌模拟题了
    ```c++
    #include <iostream>
    using namespace std;

    void op(string s)
    {
    int i, l;

    for(i = 0, l = s.size();i < l && s[i] == '0';i ++)
    ;

    //cout << endl << s << ' ' << l << ' ' << i << endl;
    /*if(s[0] == '0' && s[l - 1] != '0')
    cout << '0';
    if(i == l - 1)
    {
    if(s[i] != '0')
    cout << s[i];
    cout<<"RETY";
    return;
    }*/
    if(s[0] == '0' && (i < l && s[i] != '0'))
    cout << '0';
    if(i >= l - 1)
    {
    if(i != l)
    cout << s[i];
    return;
    }

    if(l - i >= 9)
    {
    op(s.substr(i, 1 + (l - i > 9)));
    cout << 'Y';
    op(s.substr(i + 1 + (l - i > 9), l - i - 1));
    }
    else if(l - i >= 5)
    {
    op(s.substr(i, l - i - 4));
    cout << 'W';
    op(s.substr(l - 4, 4));
    }
    else
    {
    cout << s[i];
    switch(l - i)
    {
    case 4:
    cout << 'Q';
    break;
    case 3:
    cout << 'B';
    break;
    case 2:
    cout << 'S';
    break;
    default:
    break;
    }
    op(s.substr(i + 1, l - i - 1));
    }

    return;
    }

    int main()
    {
    string s;
    int i, j, l;

    cin >> s;
    if(s[0] == '-')
    cout << 'F';
    for(i = (s[0] == '+' || s[0] == '-'), l = s.size();i < l && s[i] == '0';i ++)
    ;
    for(j = 0;j < l;j ++)
    if(s[j] == '.')
    break;

    if(i == l || s[i] == '.')
    cout << '0';
    else
    op(s.substr(i, j - i + (s[j] != '.')));

    if(s[j] == '.' && j != l - 1)
    {
    cout << 'D';
    for(j ++;j < l;j ++)
    cout << s[j];
    }
    cout << endl;

    return 0;
    }
    ```

  • 0
    @ 2016-08-13 16:09:59

    一大堆很容易把自己都绕晕啊

    Pascal Code

    var
      a,zc,xc,c4:string;
      n,nz,nx,d,i:integer;
    procedure say04(var y:string);
    begin
      if y[4]='0' then exit
                  else write(y[4]);
    end;
    procedure say03(var y:string);
    begin
      if y[3]='0' then
        if y[4]='0' then exit
                    else write('0',y[4])
                  else
                    begin
                      write(y[3],'S');
                      say04(y);
                    end;
    end;
    procedure say02(var y:string);
    begin
      if y[2]='0' then
        if y[3]='0' then
          if y[4]='0' then exit
                      else write('0',y[4])
                    else
                      begin
                        write('0',y[3],'S');
                        say04(y);
                      end
                  else
                    begin
                      write(y[2],'B');
                      say03(y);
                    end;
    end; 
    procedure say4(var y:string;w:integer);
    begin
      case w of
        4:begin
            if y[1]='0' then
              if y[2]='0' then
                if y[3]='0' then
                  if y[4]='0' then exit
                              else write('0',y[4])
                            else
                              begin
                                write('0',y[3],'S');
                                say04(y);
                              end
                          else 
                            begin
                              write('0',y[2],'B');
                              say03(y);
                            end
                        else 
                          begin
                            write(y[1],'Q');
                            say02(y);
                          end;
          end;
        3:begin
            write(y[1],'B');
            if y[2]='0' then
              if y[3]='0' then exit
                          else write('0')
                        else write(y[2],'S');
            write(y[3]);
          end;
        2:begin
            write(y[1],'S');
            if y[2]='0' then exit
                        else write(y[2]);
          end;
        1:write(y[1]);
      end;
    end;
    begin
      read(a);
      if a[1]='-' then
        begin
          write('F');
          delete(a,1,1);
        end
                  else 
                    if a[1]='+' then delete(a,1,1);
      while (a[1]='0')and(length(a)<>1) do
        delete(a,1,1);
      n:=length(a);
      if pos('.',a)<>0 then
        begin
          zc:=copy(a,1,pos('.',a)-1);
          xc:=copy(a,pos('.',a)+1,n-pos('.',a));
          nz:=length(zc);
          nx:=length(xc);
        end
                       else
                         begin
                           zc:=a;
                           nz:=n;
                         end;
      if (nz/4)<>(nz div 4) then d:=nz div 4+1
                            else d:=nz div 4;
      for i:=1 to d do
        begin
          if i=1 then
            begin
              if nz mod 4<>0 then
                begin 
                  c4:=copy(zc,1,nz mod 4);
                  say4(c4,nz mod 4);
                end
                             else 
                               begin 
                                 c4:=copy(zc,1,4);
                                 say4(c4,4);
                               end;
              if nz mod 4<>0 then delete(zc,1,nz mod 4)
                             else delete(zc,1,4);
            end
                  else
                    begin
                      c4:=copy(zc,1,4);
                      say4(c4,4);
                      delete(zc,1,4);
                    end;
            if d=3 then
              begin
                if i=1 then write('Y');
                if(i=2)and(c4='0000')and(zc[nz-4]<>'0') then write('0');
                if (i=2)and(c4<>'0000') then write('W');
              end;
          if d=2 then
            if i=1 then write('W');
        end;
      if pos('.',a)<>0 then
        if pos('.',a)=1 then write('0D',xc)
                        else if nx<>0 then write('D',xc);
    end.
    
  • 0
    @ 2016-03-29 21:47:26

    program _1230sayshu;
    var a,zc,xc,c4:string;
    n,nz,nx,d,i,kk,nkk:integer;
    procedure say04(var y:string);
    begin
    if y[4]='0' then exit
    else write(y[4]);
    end;

    procedure say03(var y:string);
    begin
    if y[3]='0' then
    if y[4]='0' then exit
    else write('0',y[4])
    else begin
    write(y[3],'S');
    say04(y);
    end;
    end;

    procedure say02(var y:string);
    begin
    if y[2]='0' then
    if y[3]='0' then
    if y[4]='0' then exit
    else write('0',y[4])
    else begin
    write('0',y[3],'S');
    say04(y);
    end
    else begin
    write(y[2],'B');
    say03(y);
    end;
    end;

    procedure say4(var y:string;w:integer);
    begin
    case w of
    4: begin
    if y[1]='0' then
    if y[2]='0' then
    if y[3]='0' then
    if y[4]='0' then exit
    else write('0',y[4])
    else begin
    write('0',y[3],'S');
    say04(y);
    end
    else begin
    write('0',y[2],'B');
    say03(y);
    end
    else begin
    write(y[1],'Q');
    say02(y);
    end;
    end;
    3: begin
    write(y[1],'B');
    if y[2]='0' then
    if y[3]='0' then exit
    else write('0')
    else write(y[2],'S');
    write(y[3]);
    end;
    2: begin
    write(y[1],'S');
    if y[2]='0' then exit
    else write(y[2]);
    end;
    1: write(y[1]);
    end;
    end;

    begin
    read(a);
    if a[1]='-' then
    begin
    write('F');
    delete(a,1,1);
    end
    else if a[1]='+' then delete(a,1,1);
    while (a[1]='0')and(length(a)<>1) do
    delete(a,1,1);
    n:=length(a);
    if pos('.',a)<>0 then
    begin
    zc:=copy(a,1,pos('.',a)-1);
    xc:=copy(a,pos('.',a)+1,n-pos('.',a));
    nz:=length(zc);
    nx:=length(xc);
    end
    else
    begin
    zc:=a;
    nz:=n;
    end;
    if nz/4<>nz div 4 then d:=nz div 4+1
    else d:=nz div 4;
    for i:=1 to d do
    begin
    if i=1 then
    begin
    if nz mod 4<>0 then
    begin c4:=copy(zc,1,nz mod 4);say4(c4,nz mod 4);end
    else begin c4:=copy(zc,1,4);say4(c4,4);end;
    if nz mod 4<>0 then delete(zc,1,nz mod 4)
    else delete(zc,1,4);
    end
    else
    begin
    c4:=copy(zc,1,4);
    say4(c4,4);
    delete(zc,1,4);
    end;
    if d=3 then
    begin
    if i=1 then write('Y');
    if(i=2)and(c4='0000')and(zc[nz-4]<>'0') then write('0');
    if (i=2)and(c4<>'0000') then write('W');
    end;
    if d=2 then
    if i=1 then write('W');
    end;
    if pos('.',a)<>0 then
    if pos('.',a)=1 then write('0D',xc)
    else if nx<>0 then write('D',xc);
    end.

  • 0
    @ 2014-07-05 23:32:08

    program _1230sayshu;
    var a,zc,xc,c4:string;
    n,nz,nx,d,i,kk,nkk:integer;
    procedure say04(var y:string);
    begin
    if y[4]='0' then exit
    else write(y[4]);
    end;

    procedure say03(var y:string);
    begin
    if y[3]='0' then
    if y[4]='0' then exit
    else write('0',y[4])
    else begin
    write(y[3],'S');
    say04(y);
    end;
    end;

    procedure say02(var y:string);
    begin
    if y[2]='0' then
    if y[3]='0' then
    if y[4]='0' then exit
    else write('0',y[4])
    else begin
    write('0',y[3],'S');
    say04(y);
    end
    else begin
    write(y[2],'B');
    say03(y);
    end;
    end;

    procedure say4(var y:string;w:integer);
    begin
    case w of
    4: begin
    if y[1]='0' then
    if y[2]='0' then
    if y[3]='0' then
    if y[4]='0' then exit
    else write('0',y[4])
    else begin
    write('0',y[3],'S');
    say04(y);
    end
    else begin
    write('0',y[2],'B');
    say03(y);
    end
    else begin
    write(y[1],'Q');
    say02(y);
    end;
    end;
    3: begin
    write(y[1],'B');
    if y[2]='0' then
    if y[3]='0' then exit
    else write('0')
    else write(y[2],'S');
    write(y[3]);
    end;
    2: begin
    write(y[1],'S');
    if y[2]='0' then exit
    else write(y[2]);
    end;
    1: write(y[1]);
    end;
    end;

    begin
    read(a);
    if a[1]='-' then
    begin
    write('F');
    delete(a,1,1);
    end
    else if a[1]='+' then delete(a,1,1);
    while (a[1]='0')and(length(a)<>1) do
    delete(a,1,1);
    n:=length(a);
    if pos('.',a)<>0 then
    begin
    zc:=copy(a,1,pos('.',a)-1);
    xc:=copy(a,pos('.',a)+1,n-pos('.',a));
    nz:=length(zc);
    nx:=length(xc);
    end
    else
    begin
    zc:=a;
    nz:=n;
    end;
    if nz/4<>nz div 4 then d:=nz div 4+1
    else d:=nz div 4;
    for i:=1 to d do
    begin
    if i=1 then
    begin
    if nz mod 4<>0 then
    begin c4:=copy(zc,1,nz mod 4);say4(c4,nz mod 4);end
    else begin c4:=copy(zc,1,4);say4(c4,4);end;
    if nz mod 4<>0 then delete(zc,1,nz mod 4)
    else delete(zc,1,4);
    end
    else
    begin
    c4:=copy(zc,1,4);
    say4(c4,4);
    delete(zc,1,4);
    end;
    if d=3 then
    begin
    if i=1 then write('Y');
    if(i=2)and(c4='0000')and(zc[nz-4]<>'0') then write('0');
    if (i=2)and(c4<>'0000') then write('W');
    end;
    if d=2 then
    if i=1 then write('W');
    end;
    if pos('.',a)<>0 then
    if pos('.',a)=1 then write('0D',xc)
    else if nx<>0 then write('D',xc);
    end.

  • 0
    @ 2014-03-20 19:16:43

    const a:array[2..10]of char=('S','B','Q','W','S','B','Q','Y','S');
    var s:string;j,ss,i,jj,sss:longint;
    begin
    readln(s);
    if s[1]='-' then begin i:=2;write('F');end else i:=1;
    if s[1]='+' then i:=2;
    if s[1]='.' then write('0');
    while(s[i]='0')and(s[i+1]<>'.')and(i<length(s))do inc(i);
    j:=i;ss:=0;
    while(s[j]<>'.')and(j<=length(s))do begin inc(j);inc(ss);end;
    while(s[i]<>'.')and(i<=length(s))do
    begin
    if s[i]='0' then
    begin
    write(0);
    while s[i]='0' do begin inc(i);dec(ss);end;
    end
    else
    begin
    if ss<>1 then
    begin
    write(s[i],a[ss]);dec(ss);jj:=i+1;sss:=0;
    if(ss=1)and(s[i+1]='0')then begin inc(i,2);break;end;
    end
    else write(s[i]);
    inc(i);
    end;
    end;
    if i>=length(s) then halt;
    write('D');inc(i);
    while i<=length(s) do begin write(s[i]);inc(i);end;
    end.

    • @ 2014-03-20 19:17:02

      抱歉 没想到 发上去就这样了 请大家用;来分句吧!!!
      sorrysorrysorrysorrysorrysorrysorrysorrysorry
      sorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorrysorry

    • @ 2014-03-20 19:17:47

      另外 60题纪念

    • @ 2014-03-20 19:18:36

      不过大牛们 你们是怎么断句的??????????????????????????????????????????

    • @ 2014-03-20 19:19:15

      补一下 ac率74%

  • 0
    @ 2013-10-30 21:54:37

    var a,i,j,k,l,m:longint;
    ans,n:string;
    b:array[1..11] of string=('','S','B','Q','W','','','','Y','','');
    bo:boolean;
    nn:double;
    begin
    readln(n);
    If n[1]='+' then Delete(N,1,1);
    if n[1]='-' then begin ans:='F';delete(n,1,1);end;
    if n[length(n)]='.' then delete(n,length(n),1);
    If n[1]='.' then n:='0'+n;
    while (n[1]='0')and(length(n)>1)and(n[2]<>'.') do delete(n,1,1);
    repeat i:=pos('.',n)-1;
    if i=-1 then i:=length(n);
    if b[i]<>'' then
    begin
    ans:=ans+n[1]+b[i];delete(n,1,1);
    end
    else
    begin
    for j:=i downto 1 do
    if b[j]<>'' then break;
    l:=i-j+1;
    ans:=ans+n[1]+b[l];
    bo:=false;
    for k:=2 to j do
    if n[k]<>'0' then begin bo:=true;break;end;
    if not(bo) then ans:=ans+b[j];
    delete(n,1,1);
    end;
    bo:=false;
    i:=pos('.',n)-1;
    if i=-1 then i:=length(n);
    for j:=1 to i do
    if n[j]<>'0' then begin bo:=true;break;end;
    if (bo)and(n[1]='0') then ans:=ans+'0';
    while (n[1]='0')and(length(n)>1) do delete(n,1,1);
    if (n[1]='.')or(n='')or(n='0') then break;
    until false;
    if n[1]='.' then
    begin
    ans:=ans+'D';
    for i:=2 to length(n) do
    ans:=ans+n[i];
    end;
    writeln(ans);
    end.

  • 0
    @ 2012-08-04 21:59:06

    别以为题目容易,其实很考验人的耐心和细心,

    注意数据会出得很极端

    (测试数据仅供参考)

    输入1:123456789

    输出1:1Y2Q3B4S5W6Q7B8S9

    输入2:1234.56789

    输出2:1Q2B3S4D56789

    输入3:+123456789.01234

    输出3:1Y2Q3B4S5W6Q7B8S9D01234

    输入4:-0000123456.0001200

    输出4:F1S2W3Q4B5S6D0001200

    输入5:-003010.

    输出5:F3Q01S

    输入6:+00101010101.00

    输出6:1Y01B01W01B01D00

    输入7:000000111111111

    输出7:1Y1Q1B1S1W1Q1B1S1

    输入8:.012

    输出8:0D012

    输入9:00000000000000

    输出9:0

    输入10:+900000001.012345600001234560000123456000012345600

    输出10:9Y01D012345600001234560000123456000012345600

    鄙人在一个多小时的艰难困苦拼命刷新AC率下,终于终于...终于AC啦!!

    #01: Accepted (28ms, 580KB)

    #02: Accepted (39ms, 580KB)

    #03: Accepted (0ms, 580KB)

    #04: Accepted (0ms, 580KB)

    #05: Accepted (0ms, 580KB)

    #06: Accepted (0ms, 580KB)

    #07: Accepted (0ms, 580KB)

    #08: Accepted (0ms, 580KB)

    #09: Accepted (0ms, 580KB)

    #10: Accepted (0ms, 580KB)

  • 0
    @ 2010-07-07 19:15:37

    program fadf;

    var i,j,n,m,k,len,p:longint; code:integer;

    s,s1,a,a1:string;

    b:array[1..15] of char;

    begin

    for i:=1 to 15 do b[i]:='p';

    b[2]:='S'; b[3]:='B'; b[4]:='Q'; b[5]:='W';

    b[6]:='S'; b[7]:='B'; b[8]:='Q'; b[9]:='Y';

    b[10]:='S'; b[11]:='B';

    readln(s);

    if s[1]='+' then delete(s,1,1)

    else if s[1]='-' then

    begin

    delete(s,1,1);

    s1:=s1+'F';

    end;

    p:=pos('.',s); if p=0 then p:=length(s)+1;

    a:=copy(s,1,p-1);

    while (length(a)>0)and(a[1]='0') do delete(a,1,1);

    if a='' then s1:=s1+'0'

    else

    begin

    for i:=1 to length(a) do

    a1:=a1+a[length(a)-i+1];

    a:=a1;

    i:=length(a);

    while i>0 do

    begin

    if a[i]'0' then

    s1:=s1+a[i]+b[i];

    dec(i); j:=0;

    while (a[i]='0')and(i>=1) do

    begin

    if i in [5,9] then

    begin

    if (a'0') or(a'0') or (a'0') then

    s1:=s1+b[i]; end;

    dec(i); inc(j);

    end;

    if (j>0)and(i>=1) then s1:=s1+'0';

    end;

    end;

    if p

  • 0
    @ 2010-07-03 08:54:11

    交了好多次终于AC了,晕死了……

    大家注意了

    第五组:-003010.

    开始没注意“.”,总是错,且不知道错在哪

  • 0
    @ 2010-04-11 21:00:57

    编译通过...

    ├ 测试数据 01:答案正确... 0ms

    ├ 测试数据 02:答案正确... 0ms

    ├ 测试数据 03:答案正确... 0ms

    ├ 测试数据 04:答案正确... 0ms

    ├ 测试数据 05:答案正确... 0ms

    ├ 测试数据 06:答案正确... 0ms

    ├ 测试数据 07:答案正确... 0ms

    ├ 测试数据 08:答案正确... 0ms

    ├ 测试数据 09:答案正确... 0ms

    ├ 测试数据 10:答案正确... 0ms

    ---|---|---|---|---|---|---|---|-

    Accepted 有效得分:100 有效耗时:0ms

    AC第50道题!!! 纪念一下。。

  • 0
    @ 2010-03-18 23:35:20

    真可恶,还有个“+”在。搞死我了

  • 0
    @ 2009-11-06 07:45:45

    试试这两个数据:

    111010101.01

    101011101.01

  • 0
    @ 2009-11-02 10:40:53

    一次AC。递归读数。

  • 0
    @ 2009-11-01 15:42:05

    谢谢楼下的提醒。

    居然有 “.012” 这样的数据。

    这种数字的写法我就没见过,更别说考虑到。结果交了N次。。。

  • 0
    @ 2009-10-14 19:04:54

    第9个点全是0 应输出0

  • 0
    @ 2009-10-08 19:54:21

    虽说只不过是简单的模拟,但是我也交了好几次才AC,模拟最重要的是细节。

    现在顿感当初学会读数字的伟大了……

  • 0
    @ 2009-10-06 16:58:01

    101010101这种数据巨猥琐啊。。

  • 0
    @ 2009-10-06 15:23:55

    第五组数据是什么啊!!!!!!!!!!!!!!!!!!!!!!!!!!??????????????

  • 0
    @ 2009-09-05 09:43:24

    编译通过...

    ├ 测试数据 01:答案正确... 0ms

    ├ 测试数据 02:答案正确... 0ms

    ├ 测试数据 03:答案正确... 0ms

    ├ 测试数据 04:答案正确... 0ms

    ├ 测试数据 05:答案正确... 0ms

    ├ 测试数据 06:答案正确... 0ms

    ├ 测试数据 07:答案正确... 0ms

    ├ 测试数据 08:答案正确... 0ms

    ├ 测试数据 09:答案正确... 0ms

    ├ 测试数据 10:答案正确... 0ms

    ---|---|---|---|---|---|---|---|-

    Accepted 有效得分:100 有效耗时:0ms

    第888个AC 以此纪念...

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