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求大神教我如何做。。。。

#1 Accepted 1ms 512.0KiB
#2 Accepted 1ms 612.0KiB
#3 Accepted 2ms 2.5MiB
#4 Accepted 2ms 2.5MiB
#5 Accepted 5ms 4.605MiB
#6 Accepted 6ms 4.5MiB
#7 Accepted 5ms 4.5MiB
#8 Accepted 4ms 4.621MiB
#9 Accepted 47ms 38.5MiB
#10 Accepted 44ms 38.5MiB
#11 Accepted 46ms 36.5MiB
#12 Accepted 43ms 36.5MiB
#13 Accepted 41ms 36.5MiB
#14 Accepted 42ms 38.5MiB
#15 Time Exceeded ≥1006ms ≥76.875MiB
#16 Time Exceeded ≥1005ms ≥79.0MiB
#17 Time Exceeded ≥1004ms ≥76.945MiB
#18 Time Exceeded ≥1004ms ≥76.875MiB
#19 Time Exceeded ≥1001ms ≥74.875MiB
#20 Time Exceeded ≥1002ms ≥79.0MiB
Var
n,num,i,j,k,max,time:longint;
t,a,sum,zh:array[1..100000]of longint;
f:array[1..10000,1..10000]of longint;
Begin
readln(n);
max:=-100;
for i:=1 to n do
begin
read(t[i],a[i]);
fillchar(sum,sizeof(sum),0);
for j:=1 to a[i] do
begin
read(f[i,j]);
if max<f[i,j] then
max:=f[i,j];
inc(zh[f[i,j]]);
end;
for j:=1 to max do
sum[j]:=zh[j];
time:=t[i]-86400;
for j:=1 to i-1 do
if time>=t[j] then
begin
for k:=1 to a[j] do
dec(sum[f[j,k]]);
end
else
break;
num:=0;
for j:=1 to max do
if sum[j]>0 then
inc(num);
writeln(num);
end;
readln;
End.

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信息

ID
2011
难度
4
分类
(无)
标签
递交数
869
已通过
235
通过率
27%
被复制
11
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