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Bài test cho các bạn tham gia tích cực tại môn Tin học .
♛๖ۣۜEɗωαɾɗ ๖ۣۜNεω๖ۣۜGαтε♛ ; Nguyễn Lê Phước Thịnh ; @Luân Trần
const fi='tvh.inp';
fo='tvh.out';
var n,d,dem,sl,s2cs,s3cs,s4cs,s5cs,s6cs,s7cs,k,i,d1:longint;
st,st1,stk:string;
f1,f2:text;
begin
assign(f1,fi); reset(f1);
assign(f2,fo); rewrite(f2);
readln(f1,n,k);
str(n,st);
d:=length(st);
case d of
1: write(9);
2: begin
sl:=n-9;
dem:=9+sl*2;
end;
3: begin
s2cs:=(99-10)+1;
s3cs:=n-99;
dem:=9+s2cs*2+s3cs*3;
end;
4: begin
s2cs:=(99-10)+1;
s3cs:=(999-100)+1;
s4cs:=n-999;
dem:=9+s2cs*2+s3cs*3+s4cs*4;
end;
5: begin
s2cs:=(99-10)+1;
s3cs:=(999-100)+1;
s4cs:=(9999-1000)+1;
s5cs:=n-9999;
dem:=9+s2cs*2+s3cs*3+s4cs*4+s5cs*5;
end;
6: begin
s2cs:=(99-10)+1;
s3cs:=(999-100)+1;
s4cs:=(9999-1000)+1;
s5cs:=(99999-10000)+1;
s6cs:=n-99999;
dem:=9+s2cs*2+s3cs*3+s4cs*4+s5cs*5+s6cs*6;
end;
7: begin
s2cs:=(99-10)+1;
s3cs:=(999-100)+1;
s4cs:=(9999-1000)+1;
s5cs:=(99999-10000)+1;
s6cs:=(999999-1000000)+1;
s7cs:=n-999999;
dem:=9+s2cs*2+s3cs*3+s4cs*4+s5cs*5+s6cs*6+s7cs*7;
end;
end;
if k<=dem then
begin
i:=1;
d1:=0;
repeat
str(i,st1);
d1:=d1+length(st1);
i:=i+1;
until d1>=k;
stk:=st1[length(st1)-(d1-k)];
writeln(f2,stk);
end;
close(f1);
close(f2);
end.
Program HOC24;
var n,i,j: integer;
s: string;
k: longint;
begin
write('Nhap N: '); readln(n);
i:=0; j:=1; k:=1;
while i<n do
begin
k:=k*j;
j:=j+1;
str(k,s);
i:=length(s);
end;
write(k);
readln
end.
a) Mình chưa hiểu vạch như thế nào , bạn ghi rõ đề giúp mình nha.
b+c)
Program hotrotinhoc;
var n:string;
max:char;
d,d1,i: integer;
begin
readln(n);
for i:=1 to length(n) do
begin
if n[i]<n[i+1] then inc(d);
if n[i]<n[i-1] then inc(d1);
if n[i]>max then max:=n[i];
end;
writeln(max);
if length(n)-1=d then write('T') else
if length(n)-1=d1 then write('L') else write(0);
readln
end.
#include <iostream>
#include <fstream>
using namespace std;
long int x[4],n,a[5001],kt[5001],ktvt[5001],MAXtong,dem=0;
int TRY(int i)
{
for(int j=x[i-1]+1;j<=n;j++)
if(kt[a[j]]==0)
{
x[i]=j;
kt[a[j]]=1;
if(i==3)
{
if(a[x[3]]==(float)(a[x[2]]+a[x[1]])/2||a[x[2]]==(float)(a[x[3]]+a[x[1]])/2||a[x[1]]==(float)(a[x[2]]+a[x[3]])/2)
{
dem++;
if(a[x[1]]+a[x[2]]+a[x[3]]>MAXtong)
{
MAXtong=a[x[1]]+a[x[2]]+a[x[3]];
}
}
}
else
TRY(i+1);
kt[a[j]]=0;
}
}
int main()
{
ifstream f("boba.inp");
f>>n;
for(int i=1;i<=n;i++)
{
f>>a[i];
}
x[0]=0;
MAXtong=-1000000000;
fill_n(kt,1001,0);
TRY(1);
cout<<dem<<endl;
if(dem>0)
{
cout<<MAXtong;
}
return 0;
}
Mình mới đạt tới trình độ quy hoạch động nên bạn thông cảm
Xin lỗi bạn, mình không hỗ trợ C. mình chỉ biết pascal thôi
const fi='tamhop.inp';
fo='tamhop.out';
var f1,f2:text;
a:array[1..100]of integer;
n,i,j,k,dem,max,t:integer;
begin
assign(f1,fi); reset(f1);
assign(f2,fo); rewrite(f2);
readln(f1,n);
for i:=1 to n do
read(f1,a[i]);
{--------------------------------xu-ly--------------------------------}
dem:=0; max:=0;
for i:=1 to n-2 do
begin
for j:=i+1 to n-1 do
begin
for k:=j+1 to n do
begin
if (a[i]=(a[j]+a[k])/2) or (a[j]=(a[i]+a[k])/2) or (a[k]=(a[i]+a[j])/2) then
begin
inc(dem);
t:=a[i]+a[j]+a[k];
if max<=t then max:=t;
end;
end;
end;
end;
writeln(f2,dem);
writeln(f2,max);
close(f1);
close(f2);
end.
const fi = 'CHIAHETK.INP';
fo = 'CHIAHETK.OUT';
MAXN = 10000;
MAXW = 1000;
oo = MAXINT div 2;
var f: text; n: integer;
k: integer;
a: array[1..MAXN] of integer;
L: array[0..MAXN, 0..MAXW] of integer;
procedure Nhap;
var i: integer;
begin
assign(f, fi); reset(f);
readln(f, n, k);
for i:= 1 to n do
read(f, a[i]);
close(f);
end;
function max(a, b: integer): integer;
begin
if a > b then exit(a)
else exit(b);
end;
function dongduduong(s: integer): integer;
begin
s:= s mod k; // -(k-1)..0..k-1
if s >= 0 then exit(s)
else exit(s + k);
end;
procedure QHD;
var i, j: integer;
begin
// Goi L[i, v] la do dai cua day con dai nhat co tong chia k du v
for j:= 0 to k-1 do
if j = dongduduong(a[1]) then L[1, j]:= 1
else L[1, j]:= -oo;
for i:= 2 to n do
for j:= 0 to k-1 do
L[i, j]:= max(1 + L[i-1, dongduduong(j-a[i])], L[i-1, j]);
end;
procedure TruyVet(i, j: integer);
begin
if i = 0 then exit;
if L[i, j] = 1 + L[i-1, dongduduong(j-a[i])] then
// Chon do vat thu i
begin
TruyVet(i-1, dongduduong(j-a[i]));
write(f, a[i], ' ');
end
else
// Khong chon do vat thu i
TruyVet(i-1, j);
end;
procedure InKQ;
var i, j:integer;
begin
assign(f, fo); rewrite(f);
writeln(f, L[n, 0]);
TruyVet(n, 0);
close(f);
end;
BEGIN
Nhap;
QHD;
InKQ;
END.
program SEQ11;
var n,k,s,dem,max:qword;
a:array[1..10000000] of qword;
i,j:longint;
begin
readln(n,k);
for i := 1 to n do read(a[i]);
max := 0; i := 1;
repeat
s := 0;
dem := 0;
for j := i to n do
begin
s := s+a[j];
inc(dem);
if s mod k = 0 then
if max < dem then max := dem;
end;
inc(i);
until i = n;
write(max);
end.
Program soa_so;
Uses crt;
Var i,j,k,l,n:longint;
st2:string;
Begin
Clrscr;
write('nhap n,k:');readln(n,k);
str(n,st2);k:=length(st2)-k;
For i:=1 to k do
Begin
l:=i;
For j:=i to length(st2)k do
If st2[l]<st2[j] then l:=j;
If l>i then delete(st2,i,l-i);
End;
Write(copy(st2,1,k));
Readln;
End.