Bai tap NMLT.pdf

BI TP MINH HA MN TIN HC C S A1

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I HC QUC GIA TPHCM TRNG H KHOA HC T NHIN

KHOA CNG NGH THNG TIN

B MN TIN HC C S

BI TP MINH HA HC PHN NHP MN LP TRNH

Mc lc:

Tun 3. CC KHI NIM C BN V K THUT LP TRNH ............................................. 2 Tun 4. CC CU TRC LP TRNH - CU TRC CHN ...................................................... 6 Tun 5. VNG LP WHILE ......................................................................................................... 11 Tun 6. VNG LP FOR .............................................................................................................. 15 Tun 7. CHNG TRNH CON ................................................................................................... 19 Tun 8. CHNG TRNH CON (tt) ............................................................................................. 20 Tun 9. KIU MNG MT CHIU V MT S K THUT C BN ................................. 22 Tun 10. TM KIM V SP XP TRN MNG MT CHIU ............................................... 27 Tun 11. MNG 2 CHIU ............................................................................................................. 32 Tun 12. 13. KIU K T V KIU CHUI ............................................................................. 39 Tun 13. QUY .......................................................................................................................... 49


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Tun 3. CC KHI NIM C BN V K THUT LP TRNH

CC BI TP C BN

Bi tp 1: Vit chng trnh in ra cc dng ch sau y: 1. In C, lowercase letters are significant.

2. main is where program execution begins.

3. Opening and closing braces enclose program statements in a routine.

4. All program statements must be terminated by a semicolon.

#include

int main (void)

{

printf ("\t1. In C, lowercase letters are significant.\n");

printf ("\t2. main is where program execution begins.\n");

printf ("\t3. Opening and closing braces enclose program statements in a

routine.\n");

printf ("\t4. All program statements must be terminated by a

semicolon.\n");

return 0;

}

Ch : 1. Ngn ng C phn bit ch hoa v ch thng. 2. Mi chng trnh lun c mt v ch mt hm main. Hm main s l ni u tin chng trnh

thc hin. 3. Mi khi c m ngoc th phi c ng ngoc. vd: {} v () 4. Cc dng lnh phi kt thc bng du chm phy ;

Bi tp 2: Vit chng trnh tnh ra kt qu ca php tr 15 cho 87, v xut kt qu ra mn hnh.

#include

int main (void)

{

int x = 15;

int y = 87;

int z = x y;

printf ("%d - %d = %d", x, y, z);

return 0;

}

Bi tp 3: Vit on chng trnh sau y m khng c cc k t bi en, sau bin dch chng trnh (F7) v xem thng bo li. Ghi ch li cc li m chng trnh thng bo. Sau , sa li chng trnh cho ng v bin dch li. Ghi ch: thng bo li s hin ra ca s pha di ca Visual C++. Nhn F4 ln lt nhy n cc li.

#include

#define TWENTYFIVE 25;

int main ()

{


Trang 3

int sum;

/* COMPUTE RESULT */

sum = TWENTYFIVE + 37 19;

/* DISPLAY RESULTS */

printf ("The answer is %i\n", sum);

return 0;

}

Bi tp 4: Vit on chng trnh sau y v d on kt qu ca chng trnh. Sau bin dch v chy chng trnh xem kt qu chnh xc. Nu kt qu khc vi mnh d on th phn tch xem ti sao li nh vy.

#include

int main ()

{

int answer, result;

answer = 100;

result = answer - 10;

printf ("The result is %i\n", result + 5);

return 0;

}

The result is 95

Bi tp 5: D on kt qu ca chng trnh sau v gii thch ti sao. Chy chng trnh v so snh kt qu tht s vi kt qu d on.

#include

#define PRINT(format,x) printf ("x = %"#format"\n", x)

int main (void)

{

int integer = 5;

char character = '5';

PRINT(d, character); PRINT(d, integer);

PRINT(c, character); PRINT(c, integer=53);

return 0;

}

x = 53

x = 5

x = 5

x = 5

Bi tp 6: D on kt qu ca chng trnh sau v gii thch ti sao. Chy chng trnh v so snh kq tht s vi kq d on.

#include

#define PR(x) printf("x = %.8g\t", (double)x)

#define PRINT4(x1,x2,x3,x4) PR(x1); PR(x2); PR(x3); PR(x4)

int main (void)

{

double d;

float f;

long l;

int i;


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i = l = f = d = 100/3; PRINT4(i, l, f, d);

i = l = f = d = 100/3. ; PRINT4(i, l, f, d);

return 0;

}

x = 33 x = 33 x = 33 x = 33 x = 33 x = 33 x = 33 x = 33.333332 x = 33.333333

Bi tp 7: Vit chng trnh tnh gi tr biu thc sau v gii thch kt qu. 3x

3 - 5x

2 + 6

Vi x = 2.55.

#include

int main (void)

{

float x = 2.55;

float y = 3*x*x*x - 5*x*x + 6;

printf ("%f", y);

return 0;

}

23.231623

CC BI TP THM C KH TRUNG BNH

1. Vit chng trnh in ln mn hnh nh sau:

*************************

* THAO CHUONG BANG *

* NGON NGU C *

*************************

2. Vit chng trnh nhp vo nm sinh, in ra tui.

V d nhp 1988 in ra: Ban sinh nam 1988 vay ban 19 tuoi.

3. Vit chng trnh thc hin cc yu cu sau (khng dng hm chuyn i):

a. Nhp vo mt k t v in ra m ASCII tng ng vi k t .

b. Nhp vo mt s nguyn (1 255) v in ra k t c m ASCII tng ng.

4. Nhp vo bn knh ca hnh trn, tnh v in ra chu vi, din tch ca hnh trn .

5. Vit chng trnh nhp vo 2 s nguyn. Xut ra min, max.

V d:

Nhp vo 5 v 7

Xut ra: min =5, max = 7

6. Tm hiu ngha cc thng bo li thng gp.

7. Tm hiu b th vin tr gip MSDN.


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CC BI TP THM C KH CAO

1. Nhp vo bn knh y R v chiu cao h ca hnh tr trn, tnh din tch y, din tch xung quanh

v th tch ca hnh tr trn theo cng thc (cc s liu l s thc, gi tr PI c C nh ngha

sn bng hng s M_PI):

a. 2RSy

b. RhSxungquanh 2

c. hSV y

2. Nhp vo s thc x, tnh v in ra cc gi tr y1, y2, ly 2 s l:

d. )1310(4 21 xxxxy

e.

xe

xxy

x

4cos

1)sin(

2

22

2

3. Nhp s tin nguyn N ng, i ra xem c bao nhiu t 10 ng, 5 ng, 2 ng v 1 ng.

V d:

N = 543 = 54 t 10 + 0 t 5 + 1 t 2 + 1 t 1

4. Nhp vo s nguyn c 3 ch s, tnh tng 3 ch s .

V d:

S 543 c tng 3 ch s l: 5 + 4 + 3 = 12

5. Vit chng trnh nhp gi, pht, giy v thc hin kim tra tnh hp l ca d liu nhp vo.

6. Vit chng trnh nhp 2 gi (gi, pht, giy) v thc hin tnh '+' v '-' ca 2 gi ny.


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Tun 4. CC CU TRC LP TRNH - CU TRC CHN

CC BI TP C BN

Bi tp 1

Vit chng trnh nhp vo mt s x, nu x = 100 th xut ra thng bo Gia tri cua x la 100,

ngc li, xut ra thng bo Gia tri ca x khac 100.

#include "stdafx.h"

#include

int main(int argc, char* argv[])

{

int x;

cout > x;

if(x == 100)

cout


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else

{

cout


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1. Gii phng trnh bc 2: ax2 + bx + c = 0 2. Nhp vo 3 s a, b, c. In ra mn hnh 3 s ny theo th t tng dn.

3. Nhp vo 4 s a, b, c, d. a. In ra s ln nht v s nh nht. b. In ra 2 s khng phi s ln nht v s nh nht.

4. Nhp vo di 3 cnh a, b, c ca 1 tam gic. a. Cho bit 3 cnh c lp thnh mt tam gic khng ? b. Nu c, cho bit loi tam gic ny (thng, cn, vung, u, vung cn).

5. Nhp 1 ch ci, nu l ch thng th i thnh ch hoa, ngc li i thnh ch thng.

6. Tnh tin i taxi t s km c nhp vo, bit:

1 km u gi 15000

T km th 2 n km th 5 gi 13500

T km th 6 tr i gi 11000

Nu i hn 120km s c gim 10% trn tng s tin.

7. Xp loi cc hc sinh trong lp. Nhp vo h tn, im ton, l, ha ca cc hc sinh. Tnh im

trung bnh 3 mn v phn loi nh sau:

sut xc: tb >=9.0

gii: 9.0 > tb >= 8.0

kh: 8.0 > tb >= 6.5

trung bnh: 6.5 > tb >= 5.0

yu: 5.0 > tb >= 3.0

km: 3.5 > tb


8. Vit chng trnh nhp vo ngy, thng, nm. Hy cho bit ngy k tip v ngy trc ca ngy . (c code tham kho bn di) #include

#include

void main()

{

//khai bao bien

int ngay, thang, nam;

int ngaytruoc, ngayke;

int nhuan;

//nhap du lieu

printf("Nhap vao mot ngay: ");

scanf("%d",&ngay);

printf("Nhap vao mot thang: ");

scanf("%d",&thang);

printf("Nhap vao mot nam: ");

scanf("%d",&nam);

//kiem tra nam nhuan

nhuan = 0;

if ((nam%400 == 0) || (nam%4 == 0 && nam%100 != 0))

nhuan = 1;


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ngaytruoc = ngay-1;

ngayke = ngay+1;

switch (thang)

{

case 1:

case 5:

case 7:

case 10:

case 12:

case 4:

case 6:

case 9:

case 11:

if (ngay == 30)

{

ngaytruoc = 29;

ngayke = 1;

}

else

{

ngaytruoc = ngay-1;

ngayke = ngay+1;

}

break;

case 2:

if (nhuan == 1)

{

if (ngay == 29)

{

ngaytruoc = 28;

ngayke = 1;

}

else if (ngay == 1)

{

ngaytruoc = 31;

ngayke = 2;

}

}

else

if (ngay == 28)

{

ngaytruoc = 27;

ngayke = 1;

}

else if (ngay == 1)

{

ngaytruoc = 31;

ngayke = 2;

}

break;

case 3:

if (ngay == 31)

{

ngaytruoc = 30;

ngayke = 1;

}

else if (ngay == 1)

{

if (nhuan == 1)

{

ngaytruoc = 29;

ngayke = 2;


Trang 10

}

else

{

ngaytruoc = 28;

ngayke = 2;

}

}

break;

case 8:

if (ngay == 31)

{

ngaytruoc = 30;

ngayke = 1;

}

else if (ngay == 1)

{

ngaytruoc = 31;

ngayke = 2;

}

break;

}

printf("Ngay truoc cua ngay %d cua thang %d cua nam %d la: %d",ngay,

thang, nam, ngaytruoc);

printf("\n");

printf("Ngay ke tiep cua ngay %d cua thang %d cua nam %d la: %d",ngay,

thang, nam, ngayke);

getch();

}


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Tun 5. VNG LP WHILE

CC BI TP C BN

Bi 1: Hy tnh tng s = 1 + 2 + 3 ..... + n

Cch 1:

#include

void main()

{

int n;

long s = 0;

printf("nhap vao n ");

scanf("%d", &n);

while (i


Trang 12

if(t==k)

{

printf("so nguyen to lon nhat nho hon %d la %d\n", n, k);

break;

}

k--;

}

if(k


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1. Kim tra xem cc ch s ca s nguyn dng n c gim dn/ tng dn t tri sang phi khng.

2. Kim tra xem 1 s nguyn dng n c phi l s i xng/ s ton s l/ s ton s chn khng

3. Tm ch s ln nht, nh nht ca s nguyn dng n.

4. 24...)1(*2*2 NNS

5. !

1

!2

1

!1

1

!0

1

nS

6. )!12(

)1...(!5!3

1253

n

xxxxS

nn vi < x <

7. )!2(

)1...(!4!2

1242

n

xxxS

nn vi < x <

8.

0

3211

1

n

nxxxxx

vi 1 < x < 1

9.

0

2642

)!2(

)1(

720

1

24

1

2

11)cos(

n

n

n

xn

xxxx vi < x <

10.

0

129753

12

)1(

29

1

7

1

5

1

3

1

2)(arccotan

n

n

n

xn

xxxxxx

11.

1

12

1

753

12

)1(

7

1

5

1

3

1)arctan(

n

n

n

xn

xxxxx vi 1 < x < 1

12.

0

432

!

1

24

1

6

1

2

11

n

nx xn

xxxxe vi < x <

13.

1

1

432 )1(

4

1

3

1

2

1)1ln(

n

n

n

xn

xxxxx vi < x <

14.

1

12753

12

2

7

2

5

2

3

22

1

1ln

n

nxn

xxxxx

x vi < x <

15.

0

12753

)!12(

)1(

5040

1

120

1

6

1)sin(

n

n

n

xn

xxxxx vi < x <

16. Nhp s nguyn dng n. (i) Kim tra co phai la s nguyn t ; (ii) Tm cc tha s nguyn

t cua no; (iii) In cac ch s t phai qua trai, (iv) In cac ch s t trai qua phai (ch s

0); (v) In ra tt ca cac cp s nguyn dng a v b (a b) sao cho: nba 22 . (vi) Nhp

s nguyn k. In ra k s thp phn ng sau du thp phn (ch : 10k c th trn s).


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17. Nhp 2 s nguyn dng a, b khc khng. (i) Tm USCLN(a, b); (ii) Tm hai s nguyn x

v y sao cho: USCLN(a, b) = a * x + b * y.

18. Nhp s nguyn dng n. Cho bit o la (i) s i xng, (ii) s gn i xng, (iii) cc ch

s xp tng dn hay giam dn khng? (iv) tng cac ch s cho n khi nho hn 10, (v) ch

s ln va nho nht.

19. Xc nh phn t th k ca dy Fibonacci: 2110

,1,0 kkk fffff , vi k 2.

20. In ra binh phng cua n s nguyn dng u tin, nhng (i) ch dng php + va phep

; (ii) Ch dng php +.


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Tun 6. VNG LP FOR

CC BI TP C BN

Bi 1: Vit chng trnh tnh tng n s t nhin u tin S = 1 + 2 + 3 + ... + n

#include "stdafx.h"

#include stdio.h

void main()

{

int n;

long S = 0;

printf(" Nhap gia tri n : ");

scanf("%d",&n);

if ( n


Trang 16

Bi 3: Vit chng trnh v mt hnh ch nht bng cc du * vi chiu di v chiu rng do ngi dng nhp vo.

#include "stdafx.h"

#include stdio.h

void main()

{

int m,n,i,j;

printf("Nhap chieu dai hinh chu nhat : m = ");

scanf("%d",&m);

printf("Nhap chieu rong hinh chu nhat : n = ");

scanf("%d",&n);

printf("\n");

for (i=0; i


Trang 17

Cch 3: #include "stdafx.h"

#include stdio.h

int x[] = {1,2,3,4,5};

int n = sizeof(x)/sizeof(int);

void main()

{

int i=0,j=n-1;

int c;

for ( ; ; ) // cau lenh FOR vang ca 3 thanh phan

{

c= x[i];

x[i] = x[j];

x[j] = c;

if (++i >= --j) break;

}

printf("\n Day ket qua la :\n");

for (i=0; i-n; ) // thay quan he i


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1. Xut tt c cc k t t a n z, A n Z, 0 n 9

2. Xut ra bng m ASCII: gm 2 ct: k t v m ASCII, yu cu hin th thnh tng trang mt,

3. Tnh tng cc s nguyn t nh hn 1000

4. Vit chng trnh nhp vo 1 s nguyn , hy vit cch c s nguyn

5. Vit chng trnh in bng cu chng ra mn hnh

6. Cn c tng 20000 t 3 loi tin 10000, 20000, 50000. Hy cho bit tt c cc phng n .

7. Cc bi ton v hnh: tam gic, hnh ch nht, cy thng,

8. Lit k tt c cc c s ca s nguyn dng n. Cho bit c bao nhiu c s v tm tng ca tt

c cc s c s .

9. Tm BSCNN ca 2 s nguyn dng a, b.

10. Kim tra 1 s c phi l s nguyn t khng.

11. Tm ch s o ngc ca s nguyn dng n

12. Tm ch s ln nht/ nh nht ca s nguyn dng n

13. m s lng ch s, tnh tng cc ch s ca s nguyn dng n

14. m s lng ch s l/ chn ca s nguyn dng n.

15. Tnh dy Fibonacci:

F0 = 0, F1 = 1, Fn = Fn-1 + Fn-2

CC BI TP THM C KH CAO Lm li cc bi kh ca chng trc vi vng lp for


Trang 19

Tun 7. CHNG TRNH CON

1. Vit hm xc nh s nh hn trong 2 s, sau s dng hm ny xc nh s nh hn

trong 3 s.

2. Vit hm tnh c s chung ln nht v bi s chung nh nht ca hai s nguyn dng a,b.

3. Vit hm tnh gi tr n! , vi n l s nguyn dng v n > 1.

n ! = 1 x 2 x ... x (n-1) x n

4. Vit hm tnh nX khng dng quy.

5. Vit chng trnh tnh hm t hp )!(!

!),(

knk

nknC

trong cn ci t hm tnh n!.

6. Vit hm tnh chu vi v din tch hnh ch nht khi bit di 2 cnh. Sau v hnh ch nht ra

mn hnh bng cc du *. Hm tnh chu vi, din tch v hm v hnh ch nht phi c lp nhau.

* * * * * * * * * * * * * * *

* *

* *

* *

* *

* * * * * * * * * * * * * * *

7. Vit chng trnh con xut ra tam gic Pascal nh sau :

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

8. Vit hm nhp vo thng bng s ri in ra tn thng bng ch ra mn hnh.

9. Vit hm kim tra mt ngy no c hp l hay khng, kim tra nm nhun.

10. Vit hm i ngy thng nm thnh th trong tun.

11. Vit hm nhn bit mt s nguyn dng c phi l s nguyn t hay khng.

12. Vit chng trnh in ra tt c cc s nguyn t nh hn s nguyn dng M cho trc ( s dng

hm kim tra s nguyn t ci t trn ).

13. Vit hm kim tra mt s nguyn dng c phi l s chnh phng hay khng. Xut tt c cc s

chnh phng trong khong A,B.


Trang 20

14. Mt s t nhin c gi l s hon thin nu n bng tng tt c cc c s ca n, k c 1. Hy

vit hm kim tra mt s c phi l s hon thin hay khng, v in ra tt c cc s hon thin nh

hn s N cho trc.

15. Vit hm tnh tng nghch o ca n s nguyn.

16. Vit hm m s cc s chn trong khong t M n N, tnh tng cc s .

17. Tnh Sin ca gi tr x bt k theo cng thc :

...!7!5!3

sin753

xxx

xx So snh kt qu vi hm sin(double) c.

18. Vit chng trnh con xut ra mn hnh dy s Fibonanci cp n, xc nh theo cng thc :

Fib(1) = 1

Fib(2) = 1

Fib(n) = Fib(n-1) + Fib(n-2) vi n> 2.

19. Ta c cc loi tin 50.000, 20.000, 10.000, 5.000, 2.000, 1.000. Vit hm cho bit s t ca tng

loi tin tng ca chng bng mt s tin no m ngi dng nhp vo. Cho bit tt c cc

phng n c th c, sau thng bit phng n no cho kt qu c s t t nht.

20. Cho trc mng s nguyn n phn t v s M. Tm tp hp cc phn t trong A sao cho tng ca

chng bng M.

Tun 8. CHNG TRNH CON (tt)

CC BI TP THM V CHNG TRNH CON

1. Vit hm i mt s h 10 sang h 16 v ngc li.

2. Vit hm lm trn mt s thc vi 2 tham s u vo : s cn phi lm trn v s ch s phn

thp phn c ngha sau khi lm trn.

3. Vit chng trnh o v tr cc k s trong mt s. D liu input l mt s nguyn dng n, gi tr

ca n s thay i sau khi gi thc hin chng trnh con o k s.

V d :

void main()

{

int n = 12345;

DaoKiSo(n);

// n == 54321

}

4. Vit chng trnh con rt gn mt phn s.

5. Vit hm tnh khong cch gia 2 im trong h ta vung gc khi bit ta ca chng.


Trang 21

6. Vit hm tnh chu vi din tch ca mt hnh ch nht, hnh tam gic trong h trc ta vung

gc khi bit ta cc nh.

7. Trong h ta -cc vung gc, cho hai im A, B c ta ln lc l (X1, Y1) v (X2, Y2) .

Vit chng trnh xc nh hai h s a,b trong phng trnh ng thng y = ax + b i qua 2 im

A, B .

8. Cho 3 im A, B, C vi cc ta tng ng ( X1, Y1) , (X2, Y2) v (X3, Y3). Vit chng trnh

xc nh trng tm ca tam gic .

9. Cho trc trong h ta cung gc cc im A, B, C v mt im X c ta bt k. Hy xc

nh xem X c nm trong tam gic hay khng.

10. Vit chng trnh in theo trt t tng dn tt c cc phn s ti gin trong khong (0,1) c mu s

khng vt qu 7.

11. Vit chng trnh con i ch thng thnh ch hoa.


Trang 22

Tun 9. KIU MNG MT CHIU V MT S K THUT C BN

CC BI TP C BN

1. Tnh tng tt c cc phn t trn mng.

#include stdafx.h

#include

#include

#include

#include

#include

void NhapMang(int[], int);

void XuatMang(int[], int);

int TinhTongCacPhanTu(int[], int);


{

int n = 10;

int a[10];

NhapMang(a, n);

XuatMang(a, n);

int s = TinhTongCacPhanTu(a, n);

cout


Trang 23

{

cout


Trang 24

void Tron2Mang(int[], int, int[], int, int[]);


{

int n = 5, m = 7;

int a[5];

int b[7];

int c[100];

srand((unsigned int)time(NULL));

cout


Trang 25


cout x;

cout


Trang 26

{

int f1 = 0;

int f2 = 1;

int f;

a[0] = f1;

a[1] = f2;

for(int i=2; i


Trang 27

Tun 10. TM KIM V SP XP TRN MNG MT CHIU

CC BI TP C BN

1. Tm mt phn t x bt k trn mng theo kiu tun t. /*********************************************************************/

/* ... */

int TimKiem(int[], int, int);


{

int n = 10;

int a[10];


NhapMang(a, n);

XuatMang(a, n);

int x;

cout > x;

int kq = TimKiem(a, n, x);

if(kq == -1)

cout


Trang 28

2. Sp xp cc phn t trn mng tng dn hoc gim dn theo yu cu. /* .................................................................. */

void SapXep(int[], int, bool);

void HoanVi(int&, int&);


{

int n = 10;

int a[100];


cout


Trang 29

3. Sp xp cc phn t trn mng sao cho cc s dng tng dn v cc s m gim dn. /*********************************************************************/

/* ... */

void SapXepDuongTangAmGiam(int a[], int n)

{

for(int i=0; i0)

{

if(a[j] < a[i])

HoanVi(a[i], a[j]);

}

if(a[j]


Trang 30

return 0;

}

/***************************************************************/

void DaoNguocMang(int a[], int n)

{

for(int i=0; i


Trang 31

5. Cho mang co n phn t. Nhp m la s nguyn dng nho hn n. Chia mang lam 2 oan a[0]

a[m 1] v a[m] a[n 1]. Khng dung thm mang phu. Chuyn ch cac phn t thanh a[m]

a[n 1] a[0] a[m 1].

6. Mng a (k phn t) v b (l phn t) cha h s cua 2 a thc. Tnh tch ca 2 a thc trn.

7. Cho 2 mng a va b co m va n phn t. Cc phn t trong mi mng l khc nhau. Tm s lng

phn t chung. M rng: gi s c phn t trng. (while)


Trang 32

Tun 11. MNG 2 CHIU

CC BI TP C BN

Bi 1: Vit chng trnh nhp, xut mt mng s nguyn hai chiu c m dng v n ct. Xc nh phn t

ln nht v nh nht trong mng.

#include stdafx.h

#include stdio.h

#define max_dong 100

#define max_cot 100

void main()

{

int m,n;

int a[max_dong][max_cot];

int i,j;

// nhap cac phan tu cho mang 2 chieu co m dong va n cot

printf(" Nhap so dong : m = ");

scanf("%d",&m);

printf("Nhap so cot : n = ");

scanf("%d",&n);

for (i=0; i


Trang 33

printf("\n Phan tu lon nhat cua mang la %d tai vi tri (%d,%d)",

max,x_max+1,y_max+1);

printf("\n Phan tu nho nhat cua mang la %d tai vi tri (%d,%d)",

min,x_min+1,y_min+1);

}

Bi 2: Vit chng trnh sp xp ma trn cc s thc tng dn t trn xung di v t tri sang phi

bng hai phng php dng v khng dng mng ph.

Cch 1 : Khng s dng mng ph.

#include stdafx.h

#include stdio.h

#define max_dong 100

#define max_cot 100

void main()

{

int m,n;


int i,j;



scanf("%d",&m);


scanf("%d",&n);

for (i=0; i


Trang 34

int i,j;



scanf("%d",&m);


scanf("%d",&n);

for (i=0; i


Trang 35

#define max_dong 50

#define max_cot 50

void main()

{

int m,n;


int i,j;



scanf("%d",&m);


scanf("%d",&n);

for (i=0; i


Trang 36


scanf("%d",&m);


scanf("%d",&n);

for (i=0; i


Trang 37

// su dung mang B luu gia tri tong cac dong cua matrix A

int b[max_dong];

int k = 0;

// tinh tong cac dong va luu vao B

for (i=0; i


Trang 38


1. Sp xp cc phn t trn mng tng dn trn tng ct v gim dn trn tng dng.

2. Sp xp cc phn t trn mng tng dn ln lut trn dng v trn ct.

3. Tnh tng v tch ca n ma trn.

4. Xoay 1 ma trn theo chiu bt k.

5. Xoay 1 ma trn theo chiu bt k n bc.

6. Xa mt hng hoc mt ct bt k trn mng 2 chiu.

7. Nhp vo 2 mng 2 chiu, tm tt c cc phn t trng nhau ca 2 mng v thay vo l s 0.

8. Nhp vo 2 ma trn cng kch thc n x m, in ra ma trn tng.

9. Nhp vo 2 ma trn cng kch thc n x m, tnh ma trn tch.

10. Nhp vo 1 ma trn, xut ra ma trn nghch o.

11. Tm gi phn t ln nht v nh nht trn tng dng, tng ct, v trn ton ma trn.


Trang 39

Tun 12. 13. KIU K T V KIU CHUI

CC BI TP C BN 1. Nhp mt chui S t bn phm. Kim tra xem chui c phi l chui i xng. V d:

Nhp: S = aBCdCBa

Xut: i xng

Nhp: S = aBCdBCa

Xut: Khng di xng

tng: Gi s chui l chui i xng. Nu pht hin 1 v tr m i xng vi n (cng v tr nhng tnh t cui chui) th chui khng cn i xng na.

Chng trnh: #include

#include

#include

// Nn nh ngha MAX c th thay i kch thc mng nhanh chng

#define MAX 100

// Nn t prototype y bit hm nhn vo ci g v tr v ci g

// u vo: Chui cn kim tra

// u ra: 1 nu l chui i xng, 0 nu l chui khng i xng

int LaChuoiDoiXung(char []);

void main()

{

char str[MAX];

printf("Nhap 1 chuoi: ");

// Nn s dng gets thay v scanf c th nhp chui c khong trng

gets(str);

int kq = LaChuoiDoiXung(str);

if (kq==1)

printf("Chuoi \"%s\" la chuoi doi xung.", str); // S dng \ xut

else

printf("Chuoi \"%s\" khong phai la chuoi doi xung.", str);

getch();

}

int LaChuoiDoiXung(char str[MAX])

{

int Flag = 1; // c gi s y l chui i xng

int l = strlen(str); // strlen ly di chui. Nn tnh 1 ln s

dng li

for (int i=0; i


Trang 40

C th khng cn s dng bin Flag kim tra m c th return 0; ngay khi kim tra thy c v tr khng cn i xng. C th nh sau: int LaChuoiDoiXung(char str[MAX])

{

int l = strlen(str); // strlen ly di chui. Nn tnh 1 ln s

dng li

for (int i=0; i


Trang 41

int KyTuXuatHienNhieuNhat(char str[], char &chr)

{

int i, j, length = strlen(str);

char curchr; // K t ang xt

int curcount; // S ln xut hin ca k t ang xt

int max = 0; // S ln xut hin nhiu nht ban u l 0 (cha c)

for (i=0; i


Trang 42

printf("Nhap ho ten: ");

// Nn s dng gets thay v scanf c th nhp chui c khong trng

gets(str);

ChuanHoaChuoi(str);

printf("Chuoi sau khi chuan hoa: %s", str);

getch();

}

void ChuanHoaChuoi(char str[])

{

int i;

// Xa cc khong trng d bn tri

// Nu k t u tin vn l khong trng th dch chuyn chui qua tri 1 k t

while (str[0]==' ')

{

for (i=0; i


Trang 43

Lu : - C th vit 1 hm xa 1 k t ti v tr x trong chui s dng chung (xem nh Bi tp) - K t kt thc chui l k t \0 - Do k t thng ng sau k t hoa 32 k t trong bng max ASCII nn mun i k t thng sang hoa th ta tr gi tr ca k t thng vi 32. n gin hn l s dng hm toupper trong th vin ctype.h.

4. Vit chng trnh nhp mt s nguyn, xut li s dng chui nhng c du , ngn cch hng triu, ngn. V d:

Nhp: N = 123456789

Xut: S = 123,456,789

tng: Chuyn s N sang chui S. i li t cui chui v c 3 k t s chn du phy vo.

Chng trnh: #include

#include

#include

#include

// Nn nh ngha MAX c th thay i kch thc mng nhanh chng

#define MAX 20


// u vo: Chui cn chn, k t chn, v tr

// u ra: khng c (chn trc tip vo chui a vo)

void ChenKyTuVaoChuoi(char [], char, int);

void main()

{

unsigned int N;

printf("Nhap mot so N: ");

scanf("%d", &N);

char str[MAX];

itoa(N, str, 10); // i s N sang chui str c s 10

for (int i=strlen(str)-3;i>0;i=i-3)

ChenKyTuVaoChuoi(str, ',', i); // Gi hm chn k t , vo chui

printf("Chuoi so sau khi xu ly la: %s", str);

getch();

}

void ChenKyTuVaoChuoi(char str[], char chr, int pos)

{

int length = strlen(str);

for (int i=length; i>pos; i--) // y chui sang phi sau v tr pos

str[i] = str[i-1];

str[pos] = chr; // a k t chr vo v tr pos

str[length+1] = '\0'; // Cp nht v tr kt thc chui

}

5. Nhp 2 chui S1 v S2 ch gm cc k s t bn phm. Xut ra tng ca 2 s .


Trang 44

V d:

Nhp: S1 = 2912 S2 = 176

Xut: S = 3088

tng 1: i 2 chui sang 2 s. Cng 2 s ri i ngc sang chui. Cch ny ch lm trong trng hp chui s ngn bin s cn sc cha.

tng 2: Thc hin php cng nh cng bng tay. tin ta thm cc s 0 vo u chui s ngn 2 chui s di bng nhau.

Chng trnh: #include

#include

#include

#define MAX 200

void main()

{

char a[MAX];

char b[MAX];

char c[MAX];

// Chui s ny lin tc nn ta s dng ngay hm scanf thay v hm gets

printf("Nhap so thu nhat: ");

scanf(%s, &a);

printf("Nhap so thu hai: ");

scanf(%s, &b);

/*

Chn s 0 vo u chui ngn hn

i vi chui ngn hn, ta thc hin 3 bc:

- o chui

- Thm s 0 vo n khi di chui ngn = chui di

- o chui li

*/

if (strlen(a)


Trang 45

temp = nc/10; // Ly phn nh

nc = nc%10;

c[clength++] = nc + 48;

}

if (temp>0)

c[clength++] = temp + 48; // Nu cn nh th a tip vo

c[clength] = '\0'; // Kt thc chui

strrev(c);

printf(" %s + %s = %s", a, b, c);

getch();

}

Nu khng thm cc s 0 vo u chui s ngn, ta c th lm nh sau: #include

#include

#include

#define MAX 200

void main()

{

char a[MAX];

char b[MAX];

char c[MAX];

int temp;

printf("Nhap so thu nhat: ");

scanf(%s, &a);

printf("Nhap so thu hai: ");

scanf(%s, &b);

strrev(a);

strrev(b);

int ablength;

ablength = strlen(a) < strlen(b) ? strlen(b) : strlen(a);

/*

Cu lnh trn tng ng vi cu lnh if sau (xem li ton t 3 ngi):

if (strlen(a) < strlen(b))

ablength = strlen(b);

else

ablength = strlen(a);

/*

int na, nb, nc, clength = 0;

temp = 0;

for (int i=0; i0)

c[clength++] = temp + 48;

c[clength] = '\0';


Trang 46

strrev(a);

strrev(b);

strrev(c);

printf("%s + %s = %s", a, b, c);

getch();

}


1. Nhp h tn ca mt ngi t bn phm. Hy chun ha chui h tn ny. (Xa cc khong trng

tha v k t u tin ca h, ch lt v tn phi vit hoa, cc k t cn li vit thng).

V du:

Nhp: NgUyen VaN A

Xut: Nguyen Van A

2. Khng s dng cc hm c sn. Vit chng trnh xa N k t ti v tr i trong chui S.

V d:

Nhp: S = Nguyen Van A i = 2 N = 3 (Xa 3 k t ti k t 2 trong chui S)

Xut: S = Nen Van A

3. Vit chng trnh nhp mt s nguyn, xut li s dng chui nhng c du , ngn cch

hng triu, ngn.

V d:

Nhp: N = 123456789

Xut: S = 123,456,789

4. Nhp mt chui S t bn phm. Kim tra xem chui c phi l chui i xng.

V d:

Nhp: S = aBCdCBa

Xut: i xng

Nhp: S = aBCdBCa

Xut: Khng di xng

5. Nhp mt chui S t bn phm. Tm k t xut hin nhiu nht trong chui v s ln xut hin.

V d:

Nhp: S = Nguyen Thi B

Xut: n 2 ln

6. Nhp mt chui S t bn phm v mt k t C. m xem k t C xut hin bao nhiu ln trong

chui S .

Nhp: Nguyen Van A C = u


Trang 47

Xut: 1 ln

7. Lp trnh nhp vo t bn phm danh sch hc sinh mt lp, sp xp li danh sch theo th t abc

ca Tn, nu trng Tn th sp xp theo th t abc ca H.

8. Vit chng trnh nhp t bn phm 2 xu k t S1 v S2. Hy xt xem S1 c xut hin bao nhiu

ln trong S2 (hoc ngc li S2 xut hin bao nhiu ln trong S1) v ti nhng v tr no?

9. Vit chng trnh nhp mt xu S ch gm cc ch ci thng. Hy lp xu S1 nhn c t xu

S bng cch sp xp li cc k t theo th t abc.

10. Cho xu S ch gm cc du ( v ). Hy kim tra xem S c l mt biu thc ( ) hp l hay

khng.

Lp trnh tnh gi tr ca mt s vit di dng LA M.

11. V d: MDCLXVI = 1666. M:1000 ; D:500 ; C:100; L:50; X :10 ; V:5 ; I :1


12. Khng s dng cc hm c sn. Vit chng trnh chn chui S2 vo chui S1 ti v tr i trong

chui S1.

V d:

Nhp: S1 = Nguyen Van A

S2 = Le

i = 8 (Chn chui S2 vo chui S1 ti v tr 8)

Xut: S1 = Nguyen LeVan A

13. Nhp 2 chui S1 v S2 ch gm cc k s t bn phm. Xut ra tng ca 2 s .

V d:

Nhp: S1 = 2912 S2 = 176

Xut: S = 3088

14. Khng s dng cc hm c sn. Kim tra xem chui S2 c nm trong chui S1 hay khng v ti v

tr no.

V d:

Nhp: S1 = Nguyen Van A S2 = Van

Xut: S2 nm trong S1 ti v tr 8.

15. Nhp mt chui S t bn phm. m xem c bao nhiu t c nhiu hn n k t c trong chui S.

Nhp: Nguyen Van A n = 2

Xut: 2 t

16. Cho xu k t S. Xt xem c hay khng mt xu X sao cho S l ghp ca mt s ln lin tip ca

X. V du S = abcabc th X l abc, cn nu S = abcab th khng c xu X.


Trang 48

17. Nhp t bn phm mt s nguyn dng N


Trang 49

Tun 13. QUY

CC BI TP C BN

1. Nhp mt s nguyn n. S dng quy tnh n! (n! = 1*2**n) V d:

Nhp: n = 0

Xut: 0! = 1

Nhp: n = 4

Xut: 4! = 24

#include

#include


// u vo: n

// u ra: n!

int GiaiThua(int);

void main()

{

int n;

printf("Nhap n: ");

scanf("%d", &n);

int kq = GiaiThua(n);

printf("%d! = %d", n, kq);

getch();

}

int GiaiThua(int n)

{

if (n==0) // Rt quan trng. y l im thot ca hm quy.

return 1;

else // C th b else v nu ri vo trng hp n=0 th return.

return n*GiaiThua(n-1);

}

Cn ch sc cha ca bin khi khai bo. Hm GiaiThua tr v kt qu kiu int c sc cha khng trong cc trng hp n ln. Lc ny s xy ra hin tng trn s v dn n cho ra kt qu sai. C th thay kiu tr v int bng kiu khc c sc cha ln hn nh float hoc double.


Trang 50

2. Nhp mt s nguyn n. S dng quy in dy Fibonacy c di n. V d:

Nhp: n = 2

Xut: 1 1

Nhp: n = 5

Xut: 1 1 2 3 5

#include

#include


// u vo: n

// u ra: phn t th n ca dy Fibonacy

int Fibonacy(int);

void main()

{

int n;

printf("Nhap n: ");

scanf("%d", &n);

for (int i=0; i


Trang 51

3. Nhp mt mng n s nguyn. S dng quy tnh tng gi tr cc phn t c trong mng. V d:

Nhp: n = 1 v mng a = 4

Xut: Tng = 4

Nhp: n = 5 v mng a = -1 4 6 3 0

Xut: Tng = 12

#include

#include

#define MAX 100

// u vo: bin cha mng cc s nguyn v s phn t

// u ra: tng cc phn t ca mng

int TongMang(int [], int);

void main()

{

int a[MAX];

for (i=0; i0)

{

printf("Mang a vua nhap la: ");

for (i=0; i


Trang 52

4. Nhp 2 s nguyn dng a v b. S dng quy tnh c s chung ln nht ca 2 s . Nhp: a = 3 b = 4

Xut: USCLN = 12

#include

#include

// u vo: 2 s nguyn dng

// u ra: c s chung ln nht ca 2 s nguyn dng

int USCLN(int, int);

void main()

{

int a, b;

printf("Nhap 2 so nguyen a va b: ");

scanf("%d%d", &a, &b);

int c = USCLN(a,b);

printf("Uoc so chung lon nhat cua %d va %d la: %d", a, b, c);

getch();

}

int USCLN(int a, int b)

{

if (a==b) // y l im thot ca hm quy a=b

return a;

if (a>b)

return USCLN(a-b, b);

return USCLN(a, b-a);

}


Trang 53

5. Nhp s nguyn dng N. S dng quy tnh in dy nh phn ca s N . Nhp: N = 5

Xut: 101

#include

#include

// u vo: s nguyn n

// u ra: dy biu din nh phn ca s nguyn dng n

void InNhiPhan(int);

void main()

{

int n;

printf("Nhap n: ");

scanf("%d", &n);

InNhiPhan(n);

getch();

}

void InNhiPhan(int n)

{

if (n/2==0) // y l im thot ca hm quy khi khng th chia n tip

printf("%d", n%2);

else // Mun b else y th phi them return; iu kin n/2==0

{

InNhiPhan(n/2); // Ch th t in: in phn cn li trc

printf("%d", n%2); // ri in s d va tm c

}

}


1. Nhp mt s nguyn N. Tnh N!

Nhp: N = 4

Xut: N! = 24

2. Nhp 2 s nguyn a v b. S dng quy tnh c s chung ln nht ca 2 s .

Nhp: a = 3 b = 4

Xut: USCLN = 12

3. Nhp mt mng gm N s nguyn. S dng quy tnh tng N s nguyn .

Nhp: [1, 5, 0, 6]

Xut: S = 12

4. Nhp mt mng gm s N s nguyn. S dng quy kim tra xem c phi l mng tng dn.

Nhp: [0, 1, 5, 6]

Xut: Tng dn

5. Nhp mt mng gm s N s nguyn. S dng quy kim tra xem c phi l mng i xng.

Nhp: [0, 1, 5, 1, 0]

Xut: Mng i xng


Trang 54

6. Nhp mt s N t bn phm. In ra s theo th t ngc li.

Nhp: N = 1234

Xut: N = 4321

7. Nhp mt s nguyn N t bn phm. Tnh gi tr biu thc:

S = 24...)1(*2*2 NN

Nhp: N = 4

Xut: S = 3,299 (S = 2468 )

8. Nhp n.

Tnh )!12(

)1...(!5!3

1253

n

xxxxS

nn .

Kim tra kt qu tm c khi n ln vi sinx

9. Nhp n. Tnh )!2(

)1...(!4!2

1242

n

xxxS

nn .

Kim tra kt qu tm c khi n ln vi cosx

10. QuickSort

11. Xy dng mt dy gm N k t t 3 k t 1, 2, 3 sao cho khng c hai dy con lin tip no ging

nhau. V d N = 5: 12321 (ng). Cc dy 12323, 12123 l khng ng.


1. Cho n s t nhin x1, x2,, xn. Hy tm UCLN (x1, x2,,xn) bng cch s dng:

UCLN(x1,x2,,xn)= UCLN(UCLN(x1,x2,,x(n-1)),xn)

2. Bi ton m i tun

3. Bi ton 8 qun hu

4. Tm tt c cc hon v ca mt mng c n phn t

5. Dy Fibonacci

6. Cho n qu cn c trng lng m1, m2,,mn. Hy tm cch t mt s qu cn sao cho cn c

cn bng.

7. Cho 1 va li c th tch V v c n vt c cc gi tr a1, a2,, an v th tch tng ng l V1,

V2,,Vn. Hy tm cch xp cc vt vo vali sao cho gi tr ca cc hng ha l cao nht

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