b:array[1..1000] of string;
c:array[1..10000] of longint;
n,i,j:longint;
s:ansistring;
begin
readln(n);
for i:=1 to n do
begin
readln(b[i]);
readln(a[i]);
end;
for i:=1 to n-1 do
for j:=n-1 downto 1 do
if (length(a[j])<length(a[j+1]))or
(length(a[j])=length(a[j+1]))and(a[j]<a[j+1])or
(a[j]=a[j+1])and(b[j]>b[j+1]) then
begin
s:=a[j];a[j]:=a[j+1];a[j+1]:=s;
s:=b[j];b[j]:=b[j+1];b[j+1]:=s;
end;
for i:=1 to n do
writeln(b[i]);
for i:=1 to n do
a[i]:=copy(a[i],length(a[i])-489,490);
for i:=1 to n do
for j:=1 to length(a[i]) do
c[j]:=c[j]+ord(a[i][length(a[i])+1-j])-48;
for i:=1 to 490 do
begin
c[i+1]:=c[i+1]+c[i] div 10;
c[i]:=c[i] mod 10;
end;
for i:=490 downto 1 do
write(c[i]);
end.
测试点5，6，7，8，9，10错了
下面是我的代码,虽然丑,但起码能AC
var
t,ans:array[1..2000] of longint;
num,nam:array[1..1000] of ansistring;
n,i,j:longint;
procedure swap(x,y:longint);
var
tmp:ansistring;
begin
tmp:=num[x];
num[x]:=num[y];
num[y]:=tmp;
tmp:=nam[x];
nam[x]:=nam[y];
nam[y]:=tmp;
end;
begin
readln(n);
for i:=1 to n do
begin
readln(nam[i]);
readln(num[i]);
end;
for i:=1 to n-1 do
for j:=i+1 to n do
if ((length(num[i])<length(num[j]))or((length(num[i])=length(num[j]))and(num[i]<num[j])))or
((num[i]=num[j])and(nam[i]>nam[j])) then swap(i,j);
for i:=1 to n do writeln(nam[i]);
for i:=1 to n do
begin
nam[i]:=copy(nam[i],length(nam[i])-489,490);
num[i]:=copy(num[i],length(num[i])-489,490);
end;
fillchar(ans,sizeof(ans),0);
for i:=1 to n do
begin
fillchar(t,sizeof(t),0);
for j:=1 to length(num[i]) do t[j]:=ord(num[i,length(num[i])+1-j])-48;
for j:=1 to 490 do
begin
inc(ans[j],t[j]);
if j<>490 then inc(ans[j+1],ans[j] div 10);
ans[j]:=ans[j] mod 10;
end;
end;
for i:=490 downto 1 do write(ans[i]);
writeln;
end.
经过思考终于找到自己的错误了
在这里也提醒一下大家
你看题目里写的是1000个猪王，即n<=1000;
数据里的n竟然有几个数据是>1000的
n估计大概<10000
所以数组最好开10000
var
i,j,m,n,k,len,la,lb,lc,x:longint;
sn,s:array[1..100000] of ansistring;
a,b,c:array[1..10000] of longint;
p:ansistring;
begin
readln(n);
for i:=1 to n do
begin
readln(sn[i]);
readln(s[i]);
end;
for i:=1 to n-1 do
for j:=i+1 to n do
if (length(s[i])<length(s[j]))
or(length(s[i])=length(s[j]))and(s[i]<s[j])
or(length(s[i])=length(s[j]))and((s[i]=s[j])and(sn[i]>sn[j])) then
begin
p:=sn[i]; sn[i]:=sn[j]; sn[j]:=p;
p:=s[i]; s[i]:=s[j]; s[j]:=p;
end;
for i:=1 to n do
begin
la:=length(s[i]);
fillchar(a,sizeof(a),0);
for j:=1 to la do a[la+1-j]:=ord(s[i][j])-ord('0');
k:=1; x:=0;
while (k<=la) or (k<=lb) do
begin
b[k]:=a[k]+b[k]+x;
x:=b[k] div 10;
b[k]:=b[k] mod 10;
inc(k);
end;
if x>0 then begin lb:=k; b[k]:=x; end
else lb:=k-1;
end;
for i:=1 to n do
writeln(sn[i]);
for i:=490 downto 1 do
write(b[i]);
end.
状态: Accepted
测评机： Xeost[5]
得分： 100分
提交日期： 2011-4-21 15:53:00
有效耗时： 3547毫秒
测试结果1: 通过本测试点|有效耗时172ms
测试结果2: 通过本测试点|有效耗时156ms
测试结果3: 通过本测试点|有效耗时250ms
测试结果4: 通过本测试点|有效耗时188ms
测试结果5: 通过本测试点|有效耗时687ms
测试结果6: 通过本测试点|有效耗时594ms
测试结果7: 通过本测试点|有效耗时63ms
测试结果8: 通过本测试点|有效耗时390ms
测试结果9: 通过本测试点|有效耗时844ms
测试结果10: 通过本测试点|有效耗时203ms
提交代码： view sourceprint?01.program jiji;
02.type num=array[0..1002] of int64;
03.var
04.a:array[0..1002] of ansistring;
05.p:array[0..1002] of num;
06.sum:num;
07.n,i,j,k,x,y,t:longint; s:ansistring; ch:char;
08.function big(x,y:integer):integer;
09.var i:longint;
10.begin
11.if p[x,0]>p[y,0] then exit(1);
12.if p[x,0]<p[y,0] then exit(2);
13.for i:=p[x,0] downto 1 do
14.begin
15.if p[x,i]>p[y,i] then exit(1);
16.if p[x,i]<p[y,i] then exit(2);
17.end;
18.exit(0);
19.end;
20.begin
21.readln(n);
22.for j:=1 to n do
23.begin
24.readln(a[j]);
25.readln(s);k:=1;t:=1;
26.for i:=length(s) downto 1 do
27.begin
28.if k=10000000 then begin k:=1;inc(t); end;
29.inc(p[j,t],k*(ord(s[i])-48));
30.k:=k*10;
31.end;
32.p[j,0]:=t;
33.end;
34.for i:=1 to n do
35.for j:=i+1 to n do
36.if (big(i,j)=2)or((big(i,j)=0) and (a[i]>a[j]))
37.then
38.begin
39.a[0]:=a[i];a[i]:=a[j];a[j]:=a[0];
40.p[0]:=p[i];p[i]:=p[j];p[j]:=p[0];
41.end;
42.for i:=1 to n do
43.for j:=1 to 70 do inc(sum[j],p[i,j]);
44.for i:=1 to 70 do begin inc(sum[i+1],sum[i] div 10000000);
45.sum[i]:=sum[i] mod 10000000;
46.end;
47.for i:=1 to n do writeln(a[i]);
48.for i:=70 downto 1 do
49.begin
50.if sum[i]<1000000 then write(0);
51.if sum[i]<100000 then write(0);
52.if sum[i]<10000 then write(0);
53.if sum[i]<1000 then write(0);
54.if sum[i]<100 then write(0);
55.if sum[i]<10 then write(0);
56.write(sum[i]);
57.end;
58.writeln;
59.end.