编译成功

Free Pascal Compiler version 2.6.4 [2014/03/06] for i386
Copyright (c) 1993-2014 by Florian Klaempfl and others
Target OS: Win32 for i386
Compiling foo.pas
foo.pas(14,7) Warning: Variable "sum" does not seem to be initialized
foo.pas(5,5) Note: Local variable "c" not used
Linking foo.exe
27 lines compiled, 0.1 sec , 28256 bytes code, 1628 bytes data
1 warning(s) issued
1 note(s) issued
测试数据 #0: RuntimeError, time = 0 ms, mem = 8640 KiB, score = 0
测试数据 #1: RuntimeError, time = 0 ms, mem = 8640 KiB, score = 0
测试数据 #2: RuntimeError, time = 15 ms, mem = 8636 KiB, score = 0
测试数据 #3: RuntimeError, time = 281 ms, mem = 8636 KiB, score = 0
测试数据 #4: TimeLimitExceeded, time = 1015 ms, mem = 8640 KiB, score = 0
测试数据 #5: TimeLimitExceeded, time = 1015 ms, mem = 8640 KiB, score = 0
测试数据 #6: TimeLimitExceeded, time = 1015 ms, mem = 8636 KiB, score = 0
测试数据 #7: WrongAnswer, time = 375 ms, mem = 8640 KiB, score = 0
测试数据 #8: WrongAnswer, time = 578 ms, mem = 8636 KiB, score = 0
测试数据 #9: WrongAnswer, time = 843 ms, mem = 8636 KiB, score = 0
TimeLimitExceeded, time = 5137 ms, mem = 8640 KiB, score = 0
代码
program nihaowoshituerbi;
var
a:array[0..10000,0..100,1..2] of longint;
b:array[0..100] of longint;
i,j,c,d,n,m,k,now,sum:longint;
begin
read(n,m);
for i:=1 to n do
for j:=0 to m-1 do
read(a[i,j,1],a[i,j,2]);
read(now);
for i:=1 to n do
begin
sum:=(sum+a[i,now,2]) mod 20123;
d:=a[i,now,2];
k:=0;
for j:=now to m-1 do
if a[i,j,1]=1 then begin inc(k);b[k]:=j;
end;
for j:=0 to now-1 do
if a[i,j,1]=1 then begin inc(k);b[k]:=j;
end;
d:=d mod k;
if d=0 then d:=k;
now:=b[d];
end;
write(sum);
end.

var
　　a,b,c:array[0..200000,1..3] of longint;
　　i,n,r,q,j:longint;
　　procedure kp(r,l:longint);
　　var
　　i,j,k1,k2:longint;
　　begin
　　i:=r;
　　j:=l;
　　k1:=a[(i+j) shr 1,1];
　　k2:=a[(i+j) shr 1,3];
　　while i<=j do begin
　　while (a[i,1]>k1) or (a[i,1]=k1) and (a[i,3]<k2) do inc(i);
　　while (a[j,1]<k1) or (a[j,1]=k1) and (a[j,3]>k2) do dec(j);
　　if i<=j then begin
　　a[0]:=a[i];
　　a[i]:=a[j];
　　a[j]:=a[0];
　　inc(i);
　　dec(j);
　　end;
　　end;
　　if i<l then kp(i,l);
　　if r<j then kp(r,j);
　　end;
　　procedure ys;
　　var
　　t,k,j:longint;
　　begin
　　i:=1;
　　j:=1;
　　t:=0;
　　while (i<n) or (j<n) do begin
　　if i>n then begin
　　for k:=j to n do begin
　　inc(t);
　　a[t]:=c[k];
　　end;
　　exit;
　　end;
　　if j>n then begin
　　for k:=i to n do begin
　　inc(t);
　　a[t]:=b[k];
　　end;
　　exit;
　　end;
　　if (b[i,1]<c[j,1]) or (b[i,1]=c[j,1]) and (b[i,3]>c[j,3]) then begin
　　inc(t);
　　a[t]:=c[j];
　　inc(j);
　　end else begin
　　inc(t);
　　a[t]:=b[i];
　　inc(i);
　　end;
　　end;
　　end;
　　begin
　　readln(n,r,q);
　　for i:=1 to n*2 do begin
　　read(a[i,1]);
　　a[i,3]:=i;
　　end;
　　for i:=1 to 2*n do read(a[i,2]);
　　kp(1,2*n);
　　for j:=1 to r do begin
　　for i:=1 to n do if a[i*2,2]>a[i*2-1,2] then begin
　　inc(a[i*2,1]);
　　b[i]:=a[i*2];
　　c[i]:=a[i*2-1];
　　end else begin
　　inc(a[i*2-1,1]);
　　b[i]:=a[i*2-1];
　　c[i]:=a[i*2];
　　end;
　　ys;
　　end;
　　writeln(a[q,3]);
　　end.

目测是道很难的题目。。。。。。。

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