508 条题解

  • 0
    @ 2008-10-26 16:53:17

    编译通过...

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    Accepted 有效得分:100 有效耗时:0ms

  • 0
    @ 2008-10-23 19:17:59

    program ex;

    var

    n,i,j,k,step,step1:longint;

    pinjun,pinyi,lunwen:longint;

    st:string;

    a:array[1..108,1..6]of string;

    b:array[1..108,1..2]of longint;

    //=====================================

    procedure init;

    begin

      readln(n);

      fillchar(b,sizeof(b),0);

      for i:=1 to n do

      begin

       readln(st);

       step:=0;

       step1:=1;

       for j:=1 to length(st) do

       if st[j]=' '

       then begin

           inc(step);

           a:=copy(st,step1,j-step1);

           step1:=j+1;

         end;

       a:=copy(st,step1,j-step1+1);

      end;

    { for i:=1 to n do

    begin

      for j:=1 to 6 do

      write(a,' ');

      writeln;

    end; }

    end;

    //======================================

    procedure jisuan;

    begin

      for i:=1 to n do

      begin

       b:=i;

       val(a,pinjun);

       val(a,pinyi);

       val(a,lunwen);

       if (pinjun>80)and(lunwen>=1)

       then inc(b,8000);

       if (pinjun>85)and(pinyi>80)

       then inc(b,4000);

       if (pinjun>90)

       then inc(b,2000);

       if (pinjun>85)and(a='Y')

       then inc(b,1000);

       if (pinyi>80)and(a='Y')

       then inc(b,850);

      end;

    end;

    //======================================

    procedure pa;

    begin

      for i:=1 to n-1 do

      for j:=i+1 to n do

      if b

  • 0
    @ 2008-10-22 19:49:05

    #include

    using namespace std;

    long n,sum,i,j,k,t,map[100]={0};

    char name[100][20],ins,sur;

    int main(){

    cin>>n;

    sum=0;

    for(i=0;i>name[i]>>j>>k>>ins>>sur>>t;

    if(j>80&&t>=1) map[i]+=8000;

    if(j>85&&k>80) map[i]+=4000;

    if(j>90) map[i]+=2000;

    if(j>85&&sur=='Y') map[i]+=1000;

    if(k>80&&ins=='Y') map[i]+=850;

    sum+=map[i];

    }

    j=0;

    for(i=1;imap[j]) j=i;

    cout

  • 0
    @ 2008-10-22 19:23:36

    type

    rec=record

    name:string;

    qm:longint;

    bj:longint;

    xg:boolean;

    xb:boolean;

    lw:boolean;

    so:longint;

    end;

    var

    r:array [1..1000] of rec;

    max:rec;

    n,i,j:integer;

    m:longint;

    str:string;

    begin

    readln(n);

    for i:=1 to n do

    begin

    readln(str);

    r[i].name:=copy(str,1,pos(' ',str)-1);

    delete(str,1,pos(' ',str));

    while str[1]' ' do begin r[i].qm:=r[i].qm*10+ord(str[1])-ord('0');delete(str,1,1);end;

    delete(str,1,1);

    while str[1]' ' do begin r[i].bj:=r[i].bj*10+ord(str[1])-ord('0');delete(str,1,1);end;

    delete(str,1,1);

    if str[1]='Y' then r[i].xg:=true else r[i].xg:=false;

    if str[3]='Y' then r[i].xb:=true else r[i].xb:=false;

    delete(str,1,4);

    if (ord(str[1])-ord('0')>=1) then r[i].lw:=true else r[i].lw:=false;

    r[i].so:=0;

    end;

    m:=0;

    max:=r[1];

    for i:=1 to n do

    begin

    r[i].so:=0;

    if (r[i].bj>80) and (r[i].xg) then inc(r[i].so,850);

    if (r[i].qm>85) and (r[i].xb) then inc(r[i].so,1000);

    if r[i].qm>90 then inc(r[i].so,2000);

    if (r[i].qm>85) and (r[i].bj>80) then inc(r[i].so,4000);

    if (r[i].qm>80) and (r[i].lw) then inc(r[i].so,8000);

    if (r[i].so)>max.so then max:=r[i];

    inc(m,r[i].so);

    end;

    writeln(max.name);

    writeln(max.so);

    writeln(m);

    end.

    用记录数组不就OK了....麻烦个啥

  • 0
    @ 2008-10-19 21:18:18

    program vp1001;

    type

    student=record

    name:string[20];

    sorce:integer;

    pi:integer;

    xg:string;

    xs:string;

    lw:integer;

    jj:integer

    end;

    var

    st:array [1..100] of student;

    i,n,max,j,t:integer;

    begin

    read(n);

    for i:=1 to n do

    begin

    readln(st[i].name);

    readln(st[i].sorce,st[i].pi);

    readln(st[i].xg);

    readln(st[i].xs);

    readln(st[i].lw)

    end;

    {with st[i] do

    begin

    readln(name);

    readln(sorce,pi);

    readln(xg);

    readln(xs);

    readln(lw)

    end; }

    for i:=1 to n do

    st[i].jj:=0;

    for i:=1 to n do

    begin

    if (st[i].sorce>80) and (st[i].lw>=1)

    then

    st[i].jj:=8000+st[i].jj;

    if (st[i].sorce>85) and (st[i].pi>80)

    then

    st[i].jj:=4000+st[i].jj;

    if st[i].sorce>90

    then

    st[i].jj:=st[i].jj+2000;

    if (st[i].sorce>85) and (st[i].xs='y')

    then

    st[i].jj:=st[i].jj+1000;

    if (st[i].pi>80) and (st[i].xs='y')

    then

    st[i].jj:=st[i].jj+850

    end;

    for i:=1 to n-1 do

    for j:=2 to n do

    if st[i].jj>st[j].jj

    then

    begin

    t:=st[j].jj;

    st[j].jj:=st[i].jj;

    st[i].jj:=t

    end;

    writeln(st[n].name);

    writeln(st[n].jj);

    max:=0;

    for i:=1 to n do

    max:=max+st[i].jj;

    writeln(max)

    end.

    真是搞不懂 如此天书式解题竟然通不过

  • 0
    @ 2008-10-19 11:52:18

    这里涉及到的就是记录不用也可以,输入数据的字符串处理,求最大值。

  • 0
    @ 2008-10-19 08:27:54

    Program P1001;

    Type rec=record

    all,name,w,c:string;

    s,e,p:integer;

    end;

    Var

    n,i,j,maxN,money,total:longint;

    cop,maxP:string;

    a:array[1..100] of rec;

    Begin

    maxN:=0;

    total:=0;

    j:=0;

    readln(n);

    For i:=1 to n do begin

    readln(a[i].all);

      while pos(' ',a[i].all)0 do begin

       j:=j+1;

       cop:=copy(a[i].all,1,pos(' ',a[i].all)-1);

       case j of

       1: a[i].name:=cop;

       2: val(cop,a[i].s);

       3: val(cop,a[i].e);

       4: a[i].c:=cop;

       5: a[i].w:=cop;

       end;

       delete(a[i].all,1,pos(' ',a[i].all));

      if pos(' ',a[i].all)=0 then val(a[i].all,a[i].p);

      end;

      j:=0;

    If (a[i].s>80) and (a[i].p>=1) then money:=8000;

    If (a[i].s>85) and (a[i].e>80) then money:=money+4000;

    If (a[i].s>90) then money:=money+2000;

    If (a[i].s>85) and ((a[i].w='Y') or (a[i].w='y')) then money:=money+1000;

    If (a[i].e>80) and ((a[i].c='Y') or (a[i].c='y')) then money:=money+850;

      If (MaxN=0) or (money>MaxN) then begin

      MaxN:=money;

      MaxP:=a[i].name;

      end;

    Total:= Total + money ;

    money:=0;

    End;

    Writeln(Maxp);

    Writeln(Maxn);

    Writeln(total);

    End.

  • 0
    @ 2008-10-15 21:16:04

    #include

    using namespace std;

    struct stu

    {

    char name[25];

    int final;

    int group;

    char leader;

    char west;

    int acd;

    int scol;

    };

    stu scolar[100];

    int main()

    {

    int n;

    cin>>n;

    stu max;

    max.scol=-1;

    double total=0;

    int i;

    for(i=0;i>scolar[i].name>>scolar[i].final>>scolar[i].group>>scolar[i].leader>>scolar[i].west>>scolar[i].acd;

    scolar[i].scol=0;

    if(scolar[i].final>80&&scolar[i].acd>=1)

    {

    scolar[i].scol+=8000;

    }

    if(scolar[i].final>85&&scolar[i].group>80)

    scolar[i].scol+=4000;

    if(scolar[i].final>90)

    scolar[i].scol+=2000;

    if(scolar[i].final>85&&scolar[i].west=='Y')

    scolar[i].scol+=1000;

    if(scolar[i].group>80&&scolar[i].leader=='Y')

    scolar[i].scol+=850;

    if(scolar[i].scol>max.scol)

    {

    max.scol=scolar[i].scol;

    strcpy(max.name,scolar[i].name);

    }

    total+=scolar[i].scol;

    }

    printf("%s\n%d\n%.0lf",max.name,max.scol,total);

    }

  • 0
    @ 2008-10-15 15:47:29

    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

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    ├ 测试数据 09:答案正确... 0ms

    ├ 测试数据 10:答案正确... 0ms

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    Accepted 有效得分:100 有效耗时:0ms

  • 0
    @ 2008-10-14 20:27:33

    注意如果最大值不唯一只输出第一个

    汗~~

    死在这里了

  • 0
    @ 2008-10-11 13:36:49

    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

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    ├ 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

  • 0
    @ 2008-10-11 11:20:51

    type rec=record

    name:string;

    m1,m2,sum:longint;

    c1,c2:char;

    m:byte;

    end;

    var

    a:array[1..100] of rec;

    i,j,k,n,ans,max:longint;

    c:char;

    begin

    readln(n);

    for i:=1 to n do begin

    read(c);

            while c' ' do begin

                a[i].name:=a[i].name+c;

                read(c)

            end;

    read(a[i].m1);

    read(a[i].m2);

    read(c);

    read(a[i].c1);

    read(c);

    read(a[i].c2);

    read(a[i].m);

    readln;

    end;

    for i:=1 to n do begin

    if (a[i].m1>80) and (a[i].m>0) then begin

    a[i].sum:=a[i].sum+8000;

    ans:=ans+8000;

    end;

    if (a[i].m1>85) and (a[i].m2>80) then begin

    a[i].sum:=a[i].sum+4000;

    ans:=ans+4000;

    end;

    if a[i].m1>90 then begin

    a[i].sum:=a[i].sum+2000;

    ans:=ans+2000;

    end;

    if (a[i].m1>85) and (a[i].c2='Y') then begin

    a[i].sum:=a[i].sum+1000;

    ans:=ans+1000;

    end;

    if (a[i].m2>80) and (a[i].c1='Y') then begin

    a[i].sum:=a[i].sum+850;

    ans:=ans+850;

    end;

    end;

    max:=0;

    for i:=1 to n do

    if a[i].sum>max then begin

    k:=i;

    max:=a[i].sum;

    end;

    writeln(a[k].name);

    writeln(max);

    writeln(ans);

    end.

  • 0
    @ 2008-10-10 21:59:15

    Program P1001;

    Type rec=record

    all,name,w,c:string;

    s,e,p:integer;

    end;

    Var

    n,i,j,maxN,money,total:longint;

    cop,maxP:string;

    a:array[1..100] of rec;

    Begin

    maxN:=0;

    total:=0;

    j:=0;

    readln(n);

    For i:=1 to n do begin

    readln(a[i].all);

    while pos(' ',a[i].all)0 do begin

    j:=j+1;

    cop:=copy(a[i].all,1,pos(' ',a[i].all)-1);

    case j of

    1: a[i].name:=cop;

    2: val(cop,a[i].s);

    3: val(cop,a[i].e);

    4: a[i].c:=cop;

    5: a[i].w:=cop;

    end;

    delete(a[i].all,1,pos(' ',a[i].all));

    if pos(' ',a[i].all)=0 then val(a[i].all,a[i].p);

    end;

    j:=0;

    If (a[i].s>80) and (a[i].p>=1) then money:=8000;

    If (a[i].s>85) and (a[i].e>80) then money:=money+4000;

    If (a[i].s>90) then money:=money+2000;

    If (a[i].s>85) and ((a[i].w='Y') or (a[i].w='y')) then money:=money+1000;

    If (a[i].e>80) and ((a[i].c='Y') or (a[i].c='y')) then money:=money+850;

    If (MaxN=0) or (money>MaxN) then begin

    MaxN:=money;

    MaxP:=a[i].name;

    end;

    Total:= Total + money ;

    money:=0;

    End;

    Writeln(Maxp);

    Writeln(Maxn);

    Writeln(total);

    End.

    好像我方法有点不一样~~~

  • 0
    @ 2008-10-06 16:51:56

    type rec=record

    name:string;

    m1,m2,sum:longint;

    c1,c2:char;

    m:byte;

    end;

    var

    a:array[1..100] of rec;

    i,j,k,n,ans,max:longint;

    c:char;

    begin

    readln(n);

    for i:=1 to n do begin

    read(c);

    while c' ' do begin

    a[i].name:=a[i].name+c;

    read(c)

    end;

    read(a[i].m1);

    read(a[i].m2);

    read(c);

    read(a[i].c1);

    read(c);

    read(a[i].c2);

    read(a[i].m);

    readln;

    end;

    for i:=1 to n do begin

    if (a[i].m1>80) and (a[i].m>0) then begin

    a[i].sum:=a[i].sum+8000;

    ans:=ans+8000;

    end;

    if (a[i].m1>85) and (a[i].m2>80) then begin

    a[i].sum:=a[i].sum+4000;

    ans:=ans+4000;

    end;

    if a[i].m1>90 then begin

    a[i].sum:=a[i].sum+2000;

    ans:=ans+2000;

    end;

    if (a[i].m1>85) and (a[i].c2='Y') then begin

    a[i].sum:=a[i].sum+1000;

    ans:=ans+1000;

    end;

    if (a[i].m2>80) and (a[i].c1='Y') then begin

    a[i].sum:=a[i].sum+850;

    ans:=ans+850;

    end;

    end;

    max:=0;

    for i:=1 to n do

    if a[i].sum>max then begin

    k:=i;

    max:=a[i].sum;

    end;

    writeln(a[k].name);

    writeln(max);

    writeln(ans);

    end.

    天啊,这水题我居然用了半个小时,和小学生一个水平,悲哀啊~~~~~~~~

  • 0
    @ 2008-10-05 21:07:42

    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

    小学生就用记录类型+字符串搞了半个小时AC。

  • 0
    @ 2008-09-29 15:11:47

    var total,i,j,k,l,m,n,b,c:longint;

    s,name:array[0..101] of string;

    money,rmb:array[0..101] of longint;

    flag1,flag2:array[0..101] of boolean;

    temp:string;

    a:longint;

    procedure qsort(l,r:longint);

    var i,j,k,x:longint;

    begin

    i:=l;j:=r;x:=money[(i+j) div 2];

    repeat

    while money[i]x do dec(j);

    if ij;

    if i80) and (a>85) then money[i]:=money[i]+4000;

    delete(s[i],1,k);

    temp:=s;

    if (temp='Y') and (b>80) then money[i]:=money[i]+850;

    delete(s[i],1,2);

    temp:=s;

    if (temp='Y') and (a>85) then money[i]:=money[i]+1000;

    delete(s[i],1,2);

    temp:=s;

    val(temp,c,l);

    if (c>=1) and (a>80) then money[i]:=money[i]+8000;

    end;

    for i:=1 to n do begin

    rmb[i]:=money[i];

    total:=total+money[i];

    end;

    qsort(1,n);

    for i:=1 to n do

    if rmb[i]=money[n] then begin

    writeln(name[i]);

    writeln(rmb[i]);

    writeln(total);

    break;

    end;

    close(input);close(output);

    end.

    快排+字符串处理

  • 0
    @ 2008-09-21 14:01:10

    #include

    int main(){

    int i,j,n,qm,py,lw,prize,max=0;

    long total=0;

    char a[20],name[20],xb,gb;

    scanf("%d",&n);

    for(i=1;i80)&&(lw>0)) prize+=8000;

    if((qm>85)&&(py>80)) prize+=4000;

    if(qm>90) prize+=2000;

    if((qm>85)&&(xb=='Y')) prize+=1000;

    if((py>80)&&(gb=='Y')) prize+=850;

    total+=prize;

    if(prize>max){

    max=prize;

    for(j=0;j

  • 0
    @ 2008-09-18 19:49:20

    program money;

    var w,f,d:char; k,len,n,i,j,temp2,sum:integer; temp1:longint;

    a:array [1..100,1..5] of longint;

    b:array [1..100,1..2] of char;

    c:array [1..100] of string;

    temp3,e:string;

    procedure se;

    begin

    if ((a>=80) and (a>0)) then a:=a+8000;

    if ((a>=85) and (a>=80)) then a:=a+4000;

    if (a>=90) then a:=a+2000;

    if ((a>=85) and ((b='Y') or (b='y'))) then a:=a+1000;

    if ((a>=80) and ((b='Y')or (b='y'))) then a:=a+850;

    end;

    begin

    fillchar(a,sizeof(a),0);

    readln(n);

    for i:=1 to n do

    begin

    repeat

    read(w);

    if w' ' then c[i]:=c[i]+w;

    until w=' ';

    readln(a,a,f,b,d,b,a);

    a:=i;

    end;

    for i:=1 to n do se;

    for i:=1 to n-1 do

    for j:=i+1 to n do

    if ((aa))) then

    begin

    temp1:=a;

    a:=a[j,4];

    a[j,4]:=temp1;

    temp2:=a;

    a:=a[j,5];

    a[j,5]:=temp2;

    temp3:=c[i];

    c[i]:=c[j];

    c[j]:=temp3;

    end;

    sum:=0;

    for i:=1 to n do sum:=sum+a;

    writeln(c[1]);

    writeln(a[1,4]);

    writeln(sum);

    end.

    为什么错误啊?

  • 0
    @ 2008-09-12 21:17:05

    1读入数据2看符合条件的人可以得到得奖学金3输出得最大奖学金的人

  • 0
    @ 2008-09-09 23:38:45

    这种水题需要的果然就是时间,去年在这卡了6次后GIVE UP,然后VIJOS就挂了。。。今年来一看。。。

    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

    5) 班级贡献奖,每人850元,班级评议成绩高于80分(>80)的学生干部均可获得;

    这里的班级评议成绩看成平均成绩了。。。靠,一道真正做了一年的题就这样解决了。

信息

ID
1001
难度
5
分类
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递交数
39260
已通过
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通过率
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