Bài tập Java
I/ Các bài tập Java cơ bản
Bài 05
/*
* I n ra m an hinh t at ca cac hop so < 100
**/
public class bai05 {
public st at ic void m ain( St ring[ ] args)
{
int k,n,dem ;
for( k= 1; k< = 100; k+ + )
{
dem = 0; / / dat j a t ri cua bien dem = 0 ung voi m oi j a t ri cua k
for( n= 2; n< = k; n+ + )
{
if( k% n= = 0) / / neu so du khi chia k cho n = 0
{
dem + + ; / / t hi bien dem duoc cong t hem 1 don vi
}
}
if( dem > 1)
{
Syst em .out .print ln( " hop so la" + k) ;
}
}
}
}
Bài 06
/*
* I n ra m an hinh 15 so nguyen t o dau t ien
**/
public class bai06 {
public st at ic void m ain( St ring[ ] args)
{
int k= 0,n,dem ,j = 0; / / khai bao cac bien, dat j a t ri ban dau cho cac bien can t hiet
while ( j < 15) / / t rong khi so nguyen t o da in < 15
{
k+ + ; / / cong cho k 1 don vi
dem = 0; / / reset gia t ri cua bien dem = 0 ung voi m oi j a t ri cua k
for ( n= 2; n< = k; n+ + )
{
if ( k% n= = 0)
{
dem + + ;
}
}
if ( dem = = 1)
{
Syst em .out .print ( k+ " " ) ;
j + + ; / / so lan in duoc cong t hem 1
1
}
}
}}
Kết quả
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47
Process com plet ed.
Bài 07
/*
* I n ra m an hinh t at ca cac so nguyen t o t u 1000 den 2000
**/
public class bai7 {
public st at ic void m ain( St ring[ ] args) {
int k,n,dem ;
for ( k= 1000; k< = 2000; k+ + )
{
dem = 0;
for ( n= 2; n< = k; n+ + )
{
if( k% n= = 0)
dem + + ;
}
if( dem = = 1)
Syst em .out .print ln( k) ;
}
}
}
Bài 08
/*
* I n ra m an hinh cac so < 100 va chia het cho 3,7
*/
public class bai08 {
public st at ic void m ain( St ring[ ] args) {
int k,n;
for ( k= 1; k< 100; k+ + )
{
if ( ( k% 3= = 0) && ( k% 7= = 0) )
Syst em .out .print ( k+ " " ) ;
}
}
}
Kết quả
21 42 63 84
Process com plet ed.
Bài 09
/*
* I n ra m an hinh cac so nam giua 1000 va 200 dong t hoi chia het cho 3,5,7
*/
public class bai09 {
2
public st at ic void m ain( St ring[ ] args) {
int k;
for ( k= 1000; k< = 2000; k+ + )
{
if( ( k% 3= = 0) &( k% 5= = 0) &( k% 7= = 0) )
Syst em .out .print ( k+ " " ) ;
}
}
}
Kết quả
1050 1155 1260 1365 1470 1575 1680 1785 1890 1995
Process com plet ed.
Bài 10
/*
* I n ra m an hinh 5 so hoan hao dau t ien( so hoan hao la so co t ong bang cac uoc so cua m inh ke ca 1)
**/
public class bai10 {
public st at ic void m ain( St ring[ ] args) {
int k= 0,j = 0,n,t ong;
while( j < 5)
{
k+ + ;
t ong= 1; / / vi 1 luon la uoc cua cac so hang
for( n= 2; n< k; n+ + ) / / j a t ri ban dau cua n = 2 t hay vi dat bang 1 vi da t inh 1 la uoc o phia t ren
{
if ( k% n= = 0)
t ong+ = n;
}
if ( k= = t ong)
{
Syst em .out .print ( k+ " " ) ;
j+ + ;
}
}
}
}
Kết quả
1 6 28 496 8128
Process com plet ed.
Bài 11
/ * Trong cac so t u nhien < = 100 hay dem xem co bao nhiu so
* - Chia het cho 5
* - Chia 5 du 1
* - Chia 5 du 2
* - Chia 5 du 3
*/
public class bai11 {
public st at ic void m ain( St ring[ ] args) {
int dem 0,dem 1,dem 2,dem 3;
int d,k;
3
dem 0= 0; dem 1= 0; dem 2= 0; dem 3= 0;
for ( k= 5; k< = 100; k + + )
{
swit ch( d= k% 5)
{
case 0 :
dem 0+ + ;
break;
case 1:
dem 1+ + ;
break;
case 2:
dem 2+ + ;
break;
case 3:
dem 3+ + ;
break;
}
}
Syst em .out .print ln( " so cac so chia het cho 5 la: " + dem 0) ;
Syst em .out .print ln( " So cac so chia 5 du 1 la: " + dem 1) ;
Syst em .out .print ln( " So cac so chia 5 du 2 la: " + dem 2) ;
Syst em .out .print ln( " So cac so chia 5 du 3 la: " + dem 3) ;
}
}
Kết quả
so cac so chia het cho 5 la: 20
So cac so chia 5 du 1 la: 19
So cac so chia 5 du 2 la: 19
So cac so chia 5 du 3 la: 19
Process com plet ed.
Bài 12
/* *
* Cho so t u nhien N bat ki( da gan t ruoc do) ,t im va in ra uoc so nguyen t o nho nhat cua N
*/
public class bai12 {
public st at ic void m ain( St ring[ ] args)
{
int N= 40;
int k;
for ( k= 2; k< = N; k+ + )
{
int dem = 0;
for( int x= 2; x< = k; x+ + )
if( k% x= = 0)
dem + + ;
if ( ( N% k= = 0) &&( dem = = 1) )
{
4
Syst em .out .print ( " uoc so nguyen t o nho nhat la: "+ k ) ;
break;
}
}
}
}
Kết quả
uoc so nguyen t o nho nhat la: 2
Process com plet ed.
Bài 13
/ * Cho so t u nhien N > 1 bat ki ( da gan t ruoc do)
* I n ra khai t rien t hanh t ich cac so nguyen t o t inh t u nho den lon
* Vd 9- - > 3.3
* 12- - > 2.2.3
*/
Cách 1 dùng for
public class bai13 {
public st at ic void m ain( St ring[ ] args) {
int N= 12;
int k;
for ( k= 2; k< = N; k+ + )
{
if ( N% k= = 0)
{
Syst em .out .print ( k+ " " ) ;
N= N/ k;
k- - ;
}
}
}
}
Cách 2: dùng while
public class bai13 {
public st at ic void m ain( St ring[ ] args) {
int N= 12,k= 2;
while( k< = N)
{
for( k= 2; k< = N; k+ + )
{
if ( N% k= = 0)
{
Syst em .out .print ( k+ " " ) ;
N= N/ k;
break;
}
}
}
}
}
Kết quả
2 2 3
5
Process com plet ed.
Bài 14
/*
* Cho t ruoc so t u nhien N bat ki ( da gan t ruoc do)
* I n ra m an hinh t at ca cac uoc so nguyen t o khac nhau cua N
*/
public class bai14 {
public st at ic void m ain( St ring[ ] args) {
int k,n,dem ,N= 1027;
for ( k= 1; k< = N; k+ + )
{
dem = 0;
for ( n= 2; n< = k; n+ + )
{
if ( k% n= = 0)
{
dem + + ;
}
}
if ( dem = = 1 & N% k= = 0)
{
Syst em .out .print ( k+ ” “ ) ;
}
}
}
}
Kết quả
13 79
Process com plet ed.
II/ Bài tập về hàm và thủ tục (Method & function)
Bài 03
/* *
* Cho so t hu nhien N bat ki
* Tinh t ong S= 1+ 1/ ( 1+ 2) + 1/ ( 1+ 2+ 3) + ... + 1/ ( 1+ 2+ 3+ ...+ N)
*/
public class ham 03 {
public st at ic void m ain( St ring[ ] args) {
int N= 2;
float S= 0;
int k;
for ( k= 1; k< = N; k+ + )
{
S+ = 1/ ( sum ( k) ) ;
}
Syst em .out .print ( " ket qua la: " + S) ;
}
public st at ic float sum ( int k)
{
int t ong= 0;
6
int x;
for ( x= 1; x< = k; x+ + )
{
t ong+ = x;
}
ret urn t ong;
}
}
Kết quả
ket qua la: 1.3333334
Process com plet ed.
Bài 04
/* *
* Cho so t u nhien N bat ki,t inh t ong
* S = 1 + 1/ 2! + 1/ 3! + ...+ 1/ N!
*/
public class ham 04 {
public st at ic void m ain( St ring[ ] args) {
int k,N= 3;
float S= 0;
for ( k= 1; k< = N; k+ + )
{
S+ = 1/ sum ( k) ;
}
Syst em .out .print ( " ket qua la: " + S) ;
}
public st at ic float sum ( int k)
{
int t ich= 1;
for ( int x= 1; x< = k; x+ + )
{
t ich= t ich* x;
}
ret urn t ich;
}
}
Kết quả
ket qua la: 1.6666666
Process com plet ed.
Bài 05
/*
* Cho so t u nhien N bat ki,t inh t ong
* S= 1 + 1/ ( 1+ 2! ) + 1/ ( 1+ 2! + 3! ) + ...+ 1/ ( 1+ 2! + 3! + ...+ N! )
*/
public class ham 05 {
public st at ic void m ain( St ring[ ] args) {
int N= 3,k;
float S= 0;
7
for ( k= 1; k< = N; k+ + )
{
S+ = 1/ sum ( k) ;
}
Syst em .out .print ( " ket qua la " + S) ;
}
public st at ic float sum ( int k)
{
float t ong= 0;
for ( int x= 1; x< = k; x+ + )
{
t ong+ = t ich( x) ;
}
ret urn t ong;
}
public st at ic float t ich( int x)
{
int t = 1;
for ( int j = 1; j < = x; j + + )
{
t= t* j;
}
ret urn t ;
}
}
Kết quả
ket qua la 1.4444445
Process com plet ed.
Bài 06
/* *
* Day Fibonaxi 1 2 3 ... F( k) = F( k- 1) + F( k- 2) .Tinh so Fibonaxi t hu N
*/
public class ham 06 {
public st at ic void m ain( St ring[ ] args) {
int a= 1,b= 2,c= 0;
int N= 10,j = 3;
while ( j < = N)
{
c= a+ b;
a= b;
b= c;
j+ + ;
}
Syst em .out .print ( " so fibonaxy t hu 10 la: " + c) ;
}
}
Kết quả
so fibonaxy t hu 10 la: 89
Process com plet ed.
8
III/ Bài tập về mảng (Array)
Bài 01
/* *
* Cho 1 day so t u nhien,viet chuong t rinh sap xep day nay t heo t hu t u giam dan.
*/
public class m ang01 {
public st at ic void m ain( St ring[ ] args) {
int [ ] a = { 3,1,7,0,10} ;
int N= 5,k,j ,t em p;
for ( k= 0; k< N- 1; k+ + )
{
for ( j = k+ 1; j < N; j + + )
{
if ( a[ k] < a[ j ] )
{
t em p= a[ j ] ;
a[ j ] = a[ k] ;
a[ k] = t em p;
}
}
}
for ( k= 0; k< N; k+ + )
Syst em .out .print ( a[ k] + " " ) ;
}
}
Kết quả:
10 7 3 1 0
Process com plet ed.
Bài 02
/* *
* Cho 1 day so t u nhien, in ra m an hinh t at ca cac so nguyen t o cua day nay
*/
public class m ang02 {
public st at ic void m ain( St ring[ ] args) {
int [ ] a = { 3,1,7,0,10} ;
int N= 5,k,x,dem ;
for ( k= 0; k< N; k+ + )
{
dem = 0;
for ( x= 2; x< = a[ k] ; x+ + )
if ( a[ k] % x= = 0)
dem + + ;
if ( dem = = 1)
Syst em .out .print ( a[ k] + " " ) ;
}
}
}
Kết quả
3 7
Process com plet ed.
9
Bài 03
/* *
* Cho 1day cac so t u nhien, t im va in ra 1 gia t ri m in cua day nay va t at ca cac chi so ung voi gt m in nay
*/
public class m ang03 {
public st at ic void m ain( St ring[ ] args) {
int [ ] a = { 3,1,7,0,10} ;
int N= 5,k,m in;
m in= a[ 0] ;
for ( k= 0; k< N; k+ + )
if ( m in > a[ k] )
m in= a[ k] ;
Syst em .out .print ln( " gia t ri nho nhat cua day la: " + m in) ;
Syst em .out .print ( " vi t ri cua so co gia t ri m in la: " ) ;
for ( k= 0; k< N; k+ + )
if ( m in = = a[ k] )
Syst em .out .print ( k+ " " ) ;
}}
Kết quả
gia t ri nho nhat cua day la: 0
vi t ri cua so co gia t ri m in la: 3
Bài 04
/* *
* Cho 1day cac so t u nhien, t im va in ra 1 gia t ri m ax cua day nay va t at ca cac chi so ung voi gt m ax nay
*/
public class m ang04 {
public st at ic void m ain( St ring[ ] args) {
int [ ] a = { 3,1,7,0,10} ;
int N= 5,k,m ax;
m ax= a[ 0] ;
for ( k= 0; k< N; k+ + )
if ( m ax < a[ k] )
m ax= a[ k] ;
Syst em .out .print ln( " gia t ri lon nhat cua day la: " + m ax) ;
Syst em .out .print ( " vi t ri cua so co gia t ri m ax la: " ) ;
for ( k= 0; k< N; k+ + )
if( m ax= = a[ k] )
Syst em .out .print ( k+ " " ) ;
}
}
Kết quả
gia t ri lon nhat cua day la: 10
vi t ri cua so co gia t ri m ax la: 4
Process com plet ed.
Bài 05
/* *
* Cho 1 day so t u nhien,hay dem xem t rong day so t ren co bao nhieu so nguyen t o, co bao nhieu hop so
*/
public class m ang05 {
public st at ic void m ain( St ring[ ] args) {
10
int
int
int
int
for
{
[ ] a = { 3,1,7,0,10} ;
N= 5,k;
nt = 0;
hs= 0;
( k= 0; k< N; k+ + )
int dem = 0;
for ( int x= 2; x< = a[ k] ; x+ + )
{
if ( a[ k] % x= = 0)
dem + + ;
}
if ( dem = = 1)
nt + + ;
else hs+ + ;
}
Syst em .out .print ln( " so cac so nguyen t o la: " + nt ) ;
Syst em .out .print ln( " so cac hop so la: " + hs) ;
}
}
Kết quả
so cac so nguyen t o la: 2
so cac hop so la: 3
Bài 06
/* *
* Cho 1 day so t u nhien,hay in ra t at ca cac so hang cua day t ren t hoa m an :
* So nay la la uoc so t huc su cua 1 so hang khac t rong day t ren
*/
public class m ang06 {
public st at ic void m ain( St ring[ ] args) {
int [ ] a = { 3,1,7,14,10} ;
int N= 5,k;
for ( k= 0; k< N; k+ + )
{
for ( int j = 0; j < N; j + + )
{
if ( ( j = = k) | ( a[ k] = = 0) )
cont inue;
if ( a[ j ] % a[ k] = = 0)
{
Syst em .out .print ( a[ k] + " " ) ;
break;
}
}
}
}
}
Kết quả
1 7
Process com plet ed.
11
Bài 07
/* *
* Cho 1 day so t u nhien,hayt im 1 so t u nhien nho nhat c khong bang bat cu so nao t rong day t ren
*/
public class m ang07 {
public st at ic void m ain( St ring[ ] args) {
int [ ] a = { 8,8,8,12,9} ;
int N= 5,k,in= 0;
for ( k= 0; k< N- 1; k+ + )
{
for ( int j = k+ 1; j < N; j + + )
{
int t em p;
if ( a[ k] > a[ j ] )
{
t em p= a[ j ] ;
a[ j ] = a[ k] ;
a[ k] = t em p;
}
}
}
for ( k= 0; k< N- 1; k+ + )
{
if( a[ k] ! = a[ k+ 1] )
{
if( k= = 0)
{
Syst em .out .print ln( a[ k] ) ;
break;
}
else
if ( a[ k- 1] ! = a[ k] )
{
Syst em .out .print ( a[ k] ) ;
break;
}
}
}
}
}
Kết quả
9
Process com plet ed.
Bài 08
/* *
* Cho 1 day so nguyen bat ki,hay xoa di t rong day nay cac so hang = 0 va in ra m an hinh cac so con lai
cua day
*/
Cách 1
public class m ang08 {
public st at ic void m ain( St ring[ ] args) {
12
int N= 8,i= 0,j = 0,dem = 0;
int [ ] a = { 8,0,0,0,0,0,12,3} ;
int [ ] b= new int [ N] ;
while( i< N)
{
if ( a[ i] = = 0)
i+ + ;
else
{
b[ j ] = a[ i] ;
i+ + ;
j+ + ;
dem + + ;
}
}
for( j = 0; j < dem ; j + + )
Syst em .out .print ( b[ j ] + " " ) ;
}
}
Cách 2
public class m ang08 {
public st at ic void m ain( St ring[ ] args) {
int [ ] a= { 8,0,0,0,12,3} ;
int N= 6,k;
int dem = 0;
for ( k= 0; k< N; k+ + )
{
if ( a[ k] = = 0)
{
dem + + ;
for ( int j = k; j < ( N- dem ) ; j + + )
a[ j ] = a[ j + 1] ;
k- - ;
}
}
for ( k= 0; k< ( N- dem ) ; k+ + )
Syst em .out .print ( a[ k] + " " ) ;
}
}
Kết quả
8 12 3
Process com plet ed.
Bài 09
/* *
* Cho 1 day o nguyen bat ki, cho t ruoc 1 so c.
* Hay dem co bao nhieu so cua day t ren = c; > c; < c.
*/
public class m ang09 {
public st at ic void m ain( St ring[ ] args) {
int N= 5,k,c= 3;
13
int [ ] a= { 10,9,8,3,5} ;
int dem 1= 0,dem 2= 0,dem 3= 0;
for ( k= 0; k< N; k+ + )
{
if ( a[ k] < c) dem 1+ + ;
if ( a[ k] = = c) dem 2+ + ;
if ( a[ k] > c) dem 3+ + ;
}
Syst em .out .print ln( " so cac so nho hon c la: " + dem 1) ;
Syst em .out .print ln( " so cac so bang c la: " + dem 2) ;
Syst em .out .print ( " so cac so lon hon c la: " + dem 3) ;
}
}
Kết quả
so cac so nho hon c la: 0
so cac so bang c la: 1
so cac so lon hon c la: 4
Process com plet ed.
Bài 10
/* *
* Cho 1 day so nguyen t o bat ki,hay t im ra 1 day so lien nhau dai nhat bao gom cac so bang nhau.
* Hay in ra so luong va cac chi so dau t ien cua day con nay
*/
public class m ang10 {
public st at ic void m ain( St ring[ ] args) {
int [ ] a= { 8,4,9,12,8,8,8,8,8} ;
int N= 9,k,dem m ax= 0,dem ,x= 0;
for ( k= 0; k< N- 1; k+ + )
{
if ( a[ k] = = a[ k+ 1] )
{
dem = 0;
for ( int j = k; j < N; j + + )
{
if ( a[ k] = = a[ j ] )
dem + + ;
if ( dem m ax< dem )
{
x= k;
dem m ax= dem ;
}
}
}
}
Syst em .out .print ln( " so cac so t huoc day dai nhat la: " + dem m ax) ;
Syst em .out .print ( " chi so cua day dai nhat la: " + x) ;
}
}
Kết quả
so cac so t huoc day dai nhat la: 5
chi so cua day dai nhat la: 4
14
Process com plet ed.
Bài 11
/* *
* Cho 1 day so nguyen bat ki. Hay t im 1 day con lien t uc don dieu t ang dai nhat cua day t ren
*/
public class m ang11 {
public st at ic void m ain( St ring[ ] args) {
int [ ] a= { 8,4,9,12,1,2,3,3,10,3} ;
int N= 10,k,dem m ax= 0,dem ,x= 0;
for ( k= 0; k< N- 1; k+ + )
{
if ( a[ k] < = a[ k+ 1] )
{
dem = 1;
for ( int j = k; j < N- 1; j + + )
{
if ( a[ j ] < = a[ j + 1] )
{
dem + + ;
if ( dem m ax< dem )
{
dem m ax= dem ;
x= k;
}
}
else break;
}
}
}
Syst em .out .print ln( " so cac so t huoc day dai nhat la: " + dem m ax) ;
Syst em .out .print ( " Chi so cua day dai nhat la: " + x) ;
}}
Kết quả
so cac so t huoc day dai nhat la: 5
Chi so cua day dai nhat la: 4
Process com plet ed.
Bài 12
/* *
* Day so a[ ] duoc goi la day con cua b[ ] neu t u b[ ] xoa di 1 vai so se t hu duoc a[ ]
* Cho t ruoc 2 day so nguyen a[ ] ; b[ ] .Hay kiem t ra xem a[ ] co la day con cua b[ ] hay ko
*/
public class m ang12 {
public st at ic void m ain( St ring[ ] args) {
int [ ] a= { 0,1,2,3} ;
int [ ] b= { 0,9,1,2,8,3,8,8,9} ;
int M= 9,N= 4,k,x= 0,j ;
int in= 0;
for( j = 0; j < N; j + + )
{
for( k= x; k< M; k+ + )
15
{
if( a[ j ] = = b[ k] )
{
in+ + ;
x= k+ 1;
break;
}
}
}
if ( in= = N)
Syst em .out .print ( " day a la day con cua day b" ) ;
else
Syst em .out .print ( " day a ko la day con cua day b" ) ;
}}
Kết quả
day a la day con cua day b
Process com plet ed.
IV/ Bài tập về xâu ( String)
Bài 01
/* *
* Cho t ruoc 1 xau ky t u la ho t en nguoi day du nhung khi nhap co t he t hua m ot so dau cach.
* Hay xoa di cac dau cach t hua va in ra ho t en chinh xac.
*/
public class St ring01 {
public st at ic void m ain( St ring[ ] args) {
St ring S = new St ring ( " Nguyen
Thi
Binh " ) ;
St ring S1,S2 = new St ring ( ) ;
S= S.t rim ( ) ;
for ( int k= 0; k< S.lengt h( ) ; k+ + )
{
S1= S.subst ring( k,k+ 1) ;
if ( S1.equals( " " ) )
{
S1= S.subst ring( k+ 1,k+ 2) ;
if ( S1.equals( " " ) )
cont inue;
else S2= S2+ S.subst ring( k,k+ 1) ;
}
else S2= S2+ S1;
}
Syst em .out .print ( S2) ;
}
}
Kết quả
Pepsi Milo Ovant ine
Process com plet ed.
Bài 02
/* *
* Cho t ruoc xau ky t u bat ky. Hay dem xem t rong xau co bao nhieu lan xuat hien xau con “ abc” .
16
*/
public class St ring02 {
public st at ic void m ain( St ring[ ] args) {
St ring S = new St ring ( " abc def ab cdfg abcabc" ) ;
St ring S1= new St ring ( ) ;
int dem = 0;
for ( int k= 0; k< S.lengt h( ) - 2; k+ + )
{
S1= S.subst ring( k,k+ 3) ;
if( S1.equals( " abc" ) )
dem + + ;
}
Syst em .out .print ( dem ) ;
}
}
Kết quả
3
Process com plet ed.
Bài 03
/* *
* Cho t ruoc 1 xau ky t u la ho t en nguoi day du, hay t ach ra phan t en cua nguoi nay.
*/
public class St ring03 {
public st at ic void m ain( St ring[ ] args) {
St ring S = new St ring ( " Nguyen Van An " ) ;
St ring S1 = new St ring ( ) ;
S= S.t rim ( ) ;
int k;
for ( k= S.lengt h( ) - 1; k> = 0; k- - )
{
S1= S.subst ring( k,k+ 1) ;
if( S1.equals( " " ) )
break;
}
Syst em .out .print ( " Ten cua nguoi do la: " + S.subst ring( k+ 1) ) ;
}
}
Kết quả
Ten cua nguoi do la: An
Process com plet ed.
Bài 04
/* *
* Cho t ruoc 1 xau ky t u la 1 ho t en nguoi day du, hay t ach ra phan ho cua nguoi nay
*/
public class St ring04 {
public st at ic void m ain( St ring[ ] args) {
St ring S = new St ring ( " Nguyen Van An " ) ;
St ring S1 = new St ring ( ) ;
S= S.t rim ( ) ;
int k;
17
for ( k= 0; k< = S.lengt h( ) ; k+ + )
{
S1= S.subst ring( k,k+ 1) ;
if( S1.equals( " " ) )
break;
}
Syst em .out .print ( " Ho cua nguoi do la: " + S.subst ring( 0,k) ) ;
}
}
Kết quả
Ho cua nguoi do la: Nguyen
Process com plet ed.
Bài 05
/* *
* Cho 1 xau ky t u bao gom t oan cac ky t u 0, 1. Hay bien doi xau nay t heo cach 0 - > 1, 1- > 0 va in ra ket
qua.
*/
public class St ring05 {
public st at ic void m ain( St ring[ ] args) {
St ring S = new St ring ( " 010001110001100" ) ;
St ring S1= new St ring ( ) ;
St ring S2= new St ring ( ) ;
for ( int k= 0; k< S.lengt h( ) ; k+ + )
{
S1= S.subst ring( k,k+ 1) ;
if ( S1.equals( " 0" ) )
S1= " 1" ;
else
S1= " 0" ;
S2= S2+ S1;
}
Syst em .out .print ( S2) ;
}
}
Kêt quả
101110001110011
Process com plet ed.
Bài 06
/* *
* Cho t ruoc xau ky t u S, in ra xau S1 nguoc lai xau S.
*/
public class St ring06 {
public st at ic void m ain( St ring[ ] args) {
St ring S,S1,S2= new St ring ( ) ;
S= " 1234567890" ;
for ( int k= S.lengt h( ) - 1; k> = 0; k- - )
{
S1= S.subst ring( k,k+ 1) ;
18
S2= S2+ S1;
}
Syst em .out .print ( S2) ;
}
}
Kết quả
0987654321
Process com plet ed.
Bài 07
/* *
* Cho t ruoc xau ky t u S. Hay bien doi S t heo quy t ac sau: chu so t hi bien t hanh “ $” con cac ky t u khac
giu nguyen.
**/
public class St ring07 {
public st at ic void m ain( St ring[ ] args) {
St ring S = new St ring ( " 12a3456b78 c 90" ) ;
St ring [ ] X= { " 0" ," 1"," 2" ," 3" ," 4","5" ," 6" ," 7" ," 8" ," 9" } ;
St ring S1= new St ring ( ) ;
St ring S2= new St ring ( ) ;
for ( int k= 0; k< S.lengt h( ) ; k+ + )
{
S1= S.subst ring( k,k+ 1) ;
for ( int j = 0; j < 10; j + + )
if ( S1.equals( X[ j ] ) )
{
S1= " $" ;
break;
}
S2= S2+ S1;
}
Syst em .out .print ( S2) ;
}
}
Kết quả
$$a$$$$b$$ c $$
Process com plet ed.
Bài 08
/* *
* Cho t ruoc 2 xau ky t u S1, S2. Hay dem xem xau S1 xuat hien t rong S2 t ai bao nhieu vi t ri.
*/
public class St ring08 {
public st at ic void m ain( St ring[ ] args) {
St ring S1= new St ring ( " abc def ghj abc ab c" ) ;
St ring S2= new St ring ( " abc" ) ;
St ring S3= new St ring( ) ;
int dem = 0;
for ( int k= 0; k< S1.lengt h( ) - S2.lengt h( ) ; k+ + )
{
S3= S1.subst ring( k,k+ S2.lengt h( ) ) ;
if ( S3.equals( S2) )
19
dem + + ;
}
Syst em .out .print ( dem ) ;
}
}
Kết quả
2
Process com plet ed.
Bài 09
/* *
* Cho xau S va 2 chi so i, j . Hay doi cho 2 vi t ri i, j t rong S.
*/
public class St ring09 {
public st at ic void m ain( St ring[ ] args) {
St ring S= new St ring ( " 0123456789" ) ;
St ring S1,S2 = new St ring( ) ;
int i= 3,j = 8;
int N= S.lengt h( ) ;
for ( int k= 0; k< N; k+ + )
{
S1= S.subst ring( k,k+ 1) ;
if ( ( k! = i) && ( k! = j ) )
S2= S2+ S1;
if ( k= = i)
S2= S2+ S.subst ring( j ,j + 1) ;
if ( k= = j )
S2= S2+ S.subst ring( i,i+ 1) ;
}
Syst em .out .print ( S2) ;
}
}
Kết quả
0128456739
Process com plet ed.
Bài 10
/* *
* Cho m ang xau ky t u S1, S2… Sn. Hay t im va in ra phan t u xau co do dai lon nhat .
*/
public class St ring10 {
public st at ic void m ain( St ring[ ] args) {
St ring [ ] S = { "Hehe" , " hahaha", " hihihihi" } ;
int m ax= 0;
for ( int k= 0; k< 3; k+ + )
{
if ( m ax< S[ k] .lengt h( ) )
m ax= S[ k] .lengt h( ) ;
}
for ( int k= 0; k< 3; k+ + )
{
if ( S[ k] .lengt h( ) = = m ax)
20
Syst em .out .print ( S[ k] ) ;
}
}
}
Kết quả
hihihihi
Process com plet ed.
Bài 11
/* *
* Cho danh sach ho t en day du hoc sinh. Hay dem xem co bao nhieu ban t en “ An” .
*/
public class St ring11 {
public st at ic void m ain( St ring[ ] args) {
St ring [ ] ds = { " Nguyen Van An " ," Nguyen Thi Binh " , " Le Van Lan " ," Le An " } ;
int dem = 0;
for( int k= 0; k< 4; k+ + )
{
ds[ k] = ds[ k] .t rim ( ) ;
int N= ds[ k] .lengt h( ) ;
St ring S1= ds[ k] .subst ring( N- 2) ;
if ( S1.equals( " An" ) )
dem + + ;
}
Syst em .out .print ( dem ) ;
}
}
Kết quả
2
Process com plet ed.
Bài 12
/* *
* Cho danh sach ho t en day du hoc sinh. Hay dem xem co bao nhieu ban co phan dem la “ Thi” .
*/
public class St ring12 {
public st at ic void m ain( St ring[ ] args) {
St ring [ ] dshs = { " Nguyen Thi Lan" , "Nguyen Thi Binh "," Le Van Lan " } ;
int dem = 0;
for ( int k= 0; k< 3; k+ + )
{
dshs[ k] = dshs[ k] .t rim ( ) ;
St ring S1= new St ring ( ) ;
int N= dshs[ k] .lengt h( ) ;
int i,j ;
for ( i= 0; i< N; i+ + )
{
S1= dshs[ k] .subst ring( i,i+ 1) ;
if( S1.equals( " " ) )
break;
21
}
for ( j = N- 1; j > = 0; j - - )
{
S1= dshs[ k] .subst ring( j ,j + 1) ;
if( S1.equals( " " ) )
break;
}
S1= dshs[ k] .subst ring( i+ 1,j ) ;
if( S1.equals( " Thi" ) )
dem + + ;
}
Syst em .out .print ( dem ) ;
}
}
Kết quả
2
Process com plet ed.
Bài 13
/* *
* Cho danh sach ho t en day du hoc sinh. Hay dem xem co bao nhieu ban co t en bat dau bang chu “ H” .
*/
Cách 1
public class St ring13 {
public st at ic void m ain( St ring[ ] args) {
St ring [ ] ds= { " Nguy en Thi Binh " ," Tran Binh Minh " ," Nguyen Thi Hoa " } ;
int i;
int dem = 0;
St ring S= new St ring( ) ;
for( int k= 0; k< 3; k+ + )
{
ds[ k] = ds[ k] .t rim ( ) ;
int N= ds[ k] .lengt h( ) ;
for ( i= N- 1; i> = 0; i- - )
{
S= ds[ k] .subst ring( i,i+ 1) ;
if ( S.equals( " " ) )
break;
}
S= ds[ k] .subst ring( i+ 1,i+ 2) ;
if( S.equals( "H" ) )
dem + + ;
}
Syst em .out .print ( dem ) ;
}
}
Cách 2
public class St ring13_2 {
public st at ic void m ain( St ring[ ] args) {
St ring [ ] ds= { " Nguy en Thi Binh " ," Tran Binh Minh " ," Nguyen Thi Hoa " } ;
22
int i;
int dem = 0;
for( int k= 0; k< 3; k+ + )
{
ds[ k] = ds[ k] .t rim ( ) ;
int N= ds[ k] .lengt h( ) ;
for ( i= N- 2; i> = 0; i- - )
{
St ring S= ds[ k] .subst ring( i,i+ 2) ;
if ( S.endsWit h( " H" ) && S.st art sWit h( " " ) )
{
dem + + ;
break;
}
}
}
Syst em .out .print ( dem ) ;
}
}
Kết quả
1
Process com plet ed.
Bài 15
/* *
* Day xau ki t u S1,S2... duoc cho t heo quy t ac sau
* S1= " 1111100000" , Sk t hu duoc t u Sk- 1 bang cach t hay doi cho lan luot cac vi t ri
* 1- 2; 2- 3; 3- 4; 4- 5; 5- 6; 6- 7; 7- 8; 8- 9; 9- 10
* Cho t ruoc so t u nhien N , hay in ra xau Sn
*/
public class St ring15 {
public st at ic void m ain( St ring[ ] args) {
St ring S = new St ring ( " 0123456789" ) ;
St ring S1 = new St ring ( ) ;
int N= 2;
int k,dem = 0;
int L= S.lengt h( ) ;
while ( dem < N)
{
for ( k= 1; k< L; k+ + )
S1= S1+ S.charAt ( k) ;
S1= S1+ S.charAt ( 0) ;
S= S1;
S1= " " ;
dem + + ;
}
Syst em .out .print ( S) ;
}
}
Kết quả
2345678901
23
Process com plet ed.
Bài 16
/* *
* Cho t ruoc 2 xau ki t u S1,S2.hay chen xau S1 vao giua xau S2 va in ra ket qua
*/
public class St ring16 {
public st at ic void m ain( St ring[ ] args) {
St ring S2= new St ring ( " 123456789" ) ;
St ring S1= new St ring( " abcdefg" ) ;
St ring S3= new St ring( ) ;
int N= S2.lengt h( ) ;
int k;
if ( N% 2= = 0) k= N/ 2;
else
k= ( N+ 1) / 2;
S3= S2.subst ring( 0,k) ;
S3= S3+ S1;
S3= S3+ S2.subst ring( k) ;
Syst em .out .print ( S3) ;
}
}
Kết quả
12345abcdefg6789
Process com plet ed.
Bài 17
/* *
* Cho t ruoc 2 xau S1,S2. Hay xet xem xau S1 o phai la xau con cua S2 neu xoa bo vai ky t u cua xau S2
duoc xau S1
*/
public class St ring17 {
public st at ic void m ain( St ring[ ] args) {
St ring S1= new St ring ( " abcdefg" ) ;
St ring S2= new St ring ( " abc3456defg789" ) ;
int x= 0,j = 0,dem = 0,k;
int N2= S2.lengt h( ) ;
int N1= S1.lengt h( ) ;
while ( j < N1)
{
k= x;
while ( k< N2)
{
if ( S2.charAt ( k) = = S1.charAt ( j ) )
{
dem + + ;
x= k;
break;
}
else k+ + ;
24
}
j+ + ;
}
if ( dem = = N1)
Syst em .out .print ( " S1 la chuoi con cua S2 " ) ;
else
Syst em .out .print ( " S1 khong phai la chuoi con cua S2" ) ;
}
}
Kết quả
S1 la chuoi con cua S2
Process com plet ed.
25
