nhap 1 mang so nguyen
a) in ra mang cac so nguyen do
b)in ra mang cac so nguyen roi sap xep theo thu tu tang dan
| m1.inp | m2.out |
| -2 3 5 4 -7 |
-2 3 5 4 -7 -7 -2 3 4 5 |
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viet phuong trinh nhap vao mang a gom N(N<=100) cac so nguyen roi thuc hien cac yeu cau
1. in ra danh sach cac so chia het cho 4
2. in ra mang hinh gia tri lon nhat chia het cho 7 va chi so do
3. xuat ra mang hinh cac phan tu cua mang do theo thu tu giam dan
uses crt;
var a,b:array[1..100]of integer;
n,i,dem,max,tam,j:integer;
begin
clrscr;
repeat
write('Nhap n='); readln(n);
until (0<n) and (n<=100);
for i:=1 to n do
begin
write('A[',i,']='); readln(a[i]);
end;
writeln('Cac so chia het cho 4 la: ');
for i:=1 to n do
if a[i] mod 4=0 then write(a[i]:4);
writeln;
dem:=0;
for i:=1 to n do
if a[i] mod 7=0 then
begin
inc(dem);
b[dem]:=a[i];
end;
max:=b[1];
for i:=1 to dem do
if max<b[i] then max:=b[i];
writeln('So lon nhat chia het cho 7 la: ',max);
writeln('Cac chi so cua no trong day A la: ');
for i:=1 to n do
if max=a[i] then write(i:4);
writeln;
for i:=1 to n-1 do
for j:=i+1 to n do
if a[i]<a[j] then
begin
tam:=a[i];
a[i]:=a[j];
a[j]:=tam;
end;
writeln('Day so sau khi sap xep giam dan la: ');
for i:=1 to n do
write(a[i]:4);
readln;
end.
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Sap xep cac phan so 1/2, 2/3, 3/4, 4/5,5/6 theo thu tu giam dan
Sap xep cac phan so 4/7, 5/9,3/4,2/5,3/2 theo thu tu tang dan
1/ Cho hinh thoi co tong hai duong cheo la 150 cm duong cheo lon hon duong cheo be 20 cm tinh dien tich hinh thoi do
sap xep cac so sau theo thu tu tang dan
43/5 ; 690% ; 6.8 ; 4 va 1/4
29/10 ; 800% ; 4.8 ; 6 va 7/10
sap xep cac so sau theo thu tu tang dan
170% ; 9.9 ; 5 ; 6 va 3/4 ; 580%
8 ; 690% ; 8.75 ; 5 va 1/4
Viet chuong trinh nhap vao 1 mang be hon 20 phan tu
a/ sap xep mang theo thu tu lon dAN
b/ gia tri trung binh cua cac phan tu
c/ in ra phan tu lon nhat va phan tu nho nhat
D/ IN Ra cac phan tu la so chan
uses crt;
var j,tg,s,max,min,i,n:longint;
a:array[1..19] of longint;
s1:real;
begin
clrscr;
write('Nhap n: '); readln(n);
for i:=1 to n do
begin
write('a[',i,']=');readln(a[i]);
end;
for i:=1 to n-1 do
for j:=i+1 to n do
if a[i]>a[j] then
begin
tg:=a[i];
a[i]:=a[j];
a[j]:=tg;
end;
for i:=1 to n do
write(a[i],' ');writeln;
for i:=1 to n do
s:=s+a[i];
s1:=s/n;
writeln(‘Gia tri trung binh cua cac phan tu la: ‘,s1:0:0);
min:=a[1];max:=a[1];
for i:=2 to n do begin
if a[i]>max then max:=a[i];
if a[i]<min then min:=a[i];
write('Phan tu lon nhat la: ',max,' ','Phan tu nho nhat la: ',min); writeln;
for i:=1 to n do
if (a[i]<>0) and (a[i] mod 2=0) then write(a[i],' ');
readln
end.
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sap xep cac so huu ti sau theo gia tri giam dan:
a.3/10;-3/4;-5/6;7/15;0
sap xep cac so huu ti sau theo gia tri tang dan:
a.3/2;-7/5;-7/9;4/5;9/11;0
b.-11/12;-3/4;-18/19;-4/5;-25/26
CAU 1:
A, Nhap vao so nguyen n ; Tinh tong S= 1+2+3+...+n
B, Nhap vao mot mang gom m so nguyen . Tinh tong cac phan tu cua mang do.
GIUP MINH VOI NHA MOI NGUOI ...
A:
uses crt;
var t,i,n:integer;
begin
clrscr;
write('Nhap n='); readln(n);
t:=0;
for i:=1 to n do
t:=t+i;
writeln(t);
readln;
end.
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B:
uses crt;
var a:array[1..100]of integer;
i,n,t:integer;
begin
clrscr;
write('Nhap m='); readln(m);
for i:=1 to m do
begin
write('A[',i,']='); readln(a[i]);
end;
t:=0;
for i:=1 to m do
t:=t+a[i];
writeln(t);
readln;
end.
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Sap xep cac phan so sau theo thu tu giam dan : 1/2 ; 2/3 ; 3/4 ; 4/5 ; 5/6 ; 6/7 ; 7/8 ; 8/9 ; 9/10
\(\frac{9}{10}>\frac{8}{9}>\frac{7}{8}>\frac{6}{7}>\frac{5}{6}>\frac{4}{5}>\frac{3}{4}>\frac{2}{3}>\frac{1}{2}.\)
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sap xep cac so huu ti sau theo thu tu tang dan
3\2;11\12;-7\4;-2\3;-7\6;11\3;5\6
2/3;5/6;11/12;7/6;3/2;7/4;11/3
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tap hop cac so nguyen x thoa man |7-x|=|x+3| la {...}.(neu tap hop co nhieu phan tu,nhap theo thu tu tang dan, cach nhau boi dau ";")
