竞赛排名 (Standard IO)

题意/Description:

 

读入/Input：

       输入文件为compe.in。文件的第一行为参赛总人数N(1<=N<=100)，从第二行到第N行依次为编号1到编号N的选手的成绩，每行有8个0～100之间的整数，代表该选手的8项竞赛成绩xi1，xi2，。。。，xi8。同一行相邻两个数之间用一个空格符隔开。

 

输出/Output：

       输出文件为compe.out。文件有N行，从第一行到第N行依次为排名第1的选手的编号，排名第2的选手的编号，...，排名第N的选手的编号。

 

题解/solution：

      只要把那一坨公式实现，然后多关键字排序就行了。不要用快排，不稳定。

 

代码/Code：

var
  a:array [1..1000,1..8] of longint;
  avg,sumy:array[1..1000] of real;
  y:array [1..1000,1..8] of real;
  sum,h:array [1..1000] of longint;
  n:longint;
procedure init;
var
  i,j:longint;
begin
  readln(n);
  for i:=1 to n do
    for j:=1 to 8 do
      read(a[i,j]);
end;

procedure try1;
var
  i,j:longint;
begin
  for i:=1 to 8 do
    begin
      for j:=1 to n do
        avg[i]:=avg[i]+a[j,i];
        avg[i]:=avg[i]/n;
    end;
end;

procedure try2;
var
  i,j:longint;
begin
  for i:=1 to n do
    begin
      for j:=1 to 8 do
        sum[i]:=sum[i]+a[i,j];
      h[i]:=i;
    end;
end;

procedure try3;
var
  i,j,k:longint;
  t:real;
begin
  for i:=1 to n do
    for j:=1 to 8 do
      begin
        t:=0;
        for k:=1 to n do
          t:=t+abs(a[k,j]-avg[j]);
        if t=0 then y[i,j]:=0
               else y[i,j]:=(a[i,j]-avg[j])*n/t;
      end;
end;

procedure try4;
var
  i,j:longint;
begin
  for i:=1 to n do
    begin
      for j:=1 to 3 do
        sumy[i]:=sumy[i]+y[i,j];
      for j:=4 to 8 do
        sumy[i]:=sumy[i]+y[i,j]*0.8;
    end;
end;

procedure main;
var
  i,j,t:longint;
  k:real;
begin
  for i:=1 to n-1 do
    for j:=i+1 to n do
      begin
        if (sumy[i]<sumy[j]) or (sumy[i]=sumy[j]) and (sum[i]<sum[j]) or
           (sumy[i]=sumy[j]) and(sum[i]=sum[j]) and (h[i]>h[j]) then
             begin
               k:=sumy[i]; sumy[i]:=sumy[j]; sumy[j]:=k;
               t:=sum[i]; sum[i]:=sum[j]; sum[j]:=t;
               t:=h[i]; h[i]:=h[j]; h[j]:=t;
             end;
      end;
end;

procedure print;
var
  i:longint;
begin
  for i:=1 to n do
    writeln(h[i]);
end;

begin
  init;
  try1;
  try2;
  try3;
  try4;
  main;
  print;
end.