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1TRNG I HC BCH KHOA H NIVIN CNG NGH THNG TIN V TRUYN THNG

TIN HC I CNGBi 9. Mng v xu k t

Ni dung

9.1. Mng

9.2. Xu k t

2

Ni dung

9.1. Mng

9.1.1. Khi nim mng

9.1.2. Khai bo v s dng mng

9.1.3. Cc thao tc c bn trn mng

9.1.4. Tm kim trn mng

9.1.5. Sp xp trn mng

9.2. Xu k t

3

9.1.1. Khi nim mng

Tp hp hu hn cc phn t cng kiu, lu tr k tip nhau trong b nh

Cc phn t trong mng c cng tn (l tn mng) nhng phn bit vi nhau ch s cho bit v tr ca n trong mng

V d:

Bng im ca sinh vin

Vector

Ma trn

4


Khai bo mng (mt chiu)kiu_d_liu tn_mng [kch_thc_mng];

Trong

kiu_d_liu: kiu d liu ca cc phn t trong mng

tn_mng: tn ca mng

kch_thc_mng: s phn t trong mng

V d

int mang_nguyen[10]; // khai bo mng 10 phn t c kiu d liu int

5


Cp pht b nh

Cc phn t trong mng c cp pht cc nh k tip nhau trong b nh

Bin mng lu tr a ch nh u tin trong vng nh c cp pht

Ngn ng C nh ch s cc phn t trong mng bt u t 0

Phn t th i trong mang_nguyen c xc nh bi mang_nguyen[i-1]

6

mang_nguyen[0] mang_nguyen[1] .. mang_nguyen[9]

mang_nguyen


Mng mt chiu v mng nhiu chiu

Mi phn t ca mng cng l mt mng

=> mng nhiu chiu

V d

int a[6][5] ; //mng 2 chiu

int b[3][4][5]; // mng 3 chiu

7


Khai bo mt bin mng

kiu_d_liu tn_mng[size1][size2][sizek];

Trong

sizei l kch thc chiu th i ca mng

8


S dng mng

Truy cp vo phn t thng qua tn mng v ch s ca phn t trong mng

tn_mng[ch_s_phn_t]

Ch : ch s bt u t 0

V d

int a[4];

phn t u tin (th nht) ca mng: a[0]

phn t cui cng (th t) ca mng: a[3]

a[i]: l phn t th i+1 ca a9


V d (tip)

int b[3][4];

phn t u tin ca mng: b[0] l mt mng mt chiu

phn t u tin ca mng b[0]: b[0][0]

b[i][j]: l phn t th j+1 ca b[i], b[i] l phn t th i+1 ca b

10


a. Nhp d liu cho mng

Khi to gi tr cho mng ngay khi khai bo

Ging vi?

int a[4] = {1,4,6,2};

int b[2][3]={ {1,2,3}, {4,5,6} }:

S lng gi tr khi to khng c ln hn s lng phn t trong mng

Nu s lng ny nh hn, cc phn t cn li c khi to gi tr 0

11



C th xc nh kch thc mng thng qua s gi tr khi to nu trng kch thc mng

int array1 [8] = {2, 4, 6, 8, 10, 12, 14, 16};

int array2 [] = {2, 4, 6, 8, 10, 12, 14, 16};

12



Nhp d liu t bn phm bng hm scanf

int a[10];

Nhp d liu cho a[1]: scanf(%d, & a[1]);

Nhp d liu cho ton b phn t ca mng a

=> S dng vng lp for

Lu

Tn mng l mt hng (hng con tr) do khng th thc hin php ton vi tn mng nh php gn sau khi khai bo

13


#include #define MONTHS 12int main(){

int rainfall[MONTHS], i;for ( i=0; i < MONTHS; i++ ){

printf(Nhap vao phan tu thu %d: , i+1);

scanf("%d", &rainfall[i] );} return 0;

}

14



Lu

Nu s phn t ca mng c nhp t bn phm v ch bit trc s phn t ti a ti a => khai bo mng vi kch thc ti a v s dng bin lu s phn t thc s ca mng.

V d: Khai bo mng s nguyn a c ti a 100 phn t. Nhp t bn phm s phn t trong mng v gi tr cc phn t .

15


#include

#include

void main(){

int a[100];

int n, i;

do{

printf(\n Cho biet so phan tu cua mang: );

scanf(%d,&n);

}while (n>100||n


b. Xut d liu trong mng

Dng hm printf()

hin th tt c cc phn t: dng vng for

V d

Hin th mt phn t bt k

Hin th tt c cc phn t, mi phn t trn mt dng

Hin th tt c cc phn t trn mt dng, cch nhau 2 v tr

Hin th tng k phn t trn mt dng

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#include #define MONTHS 12int main(){

int rainfall[MONTHS], i;for ( i=0; i < MONTHS; i++ ){

printf(Nhap vao phan tu thu %d: , i+1);

scanf("%d", &rainfall[i] );} for ( i=0; i < MONTHS; i++ )

printf( "%2d , rainfall[i]);printf("\n");

return 0;}

18


c. Tm gi tr ln nht, nh nht

Tm gi tr ln nht

Gi s phn t l phn t u tin

Ln lt so snh vi cc phn t cn li

Nu ln hn hoc bng => so snh tip

Nu nh hn => coi phn t ny l phn t ln nht v tip tc so snh

Cch lm?

Tm gi tr nh nht: tng t

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max = rainfall[0];

for(i = 1; i < n; i++)

if(max < a[i])

max = a[i];

printf("\n Luong mua nhieu nhat la: %d", max);

20


Bi ton

Cho mng d liu a v mt gi tr k

Tm cc phn t trong mng a c gi tr bng (ging) vi k. Nu c in ra v tr (ch s) cc phn t ny. Ngc li thng bo khng tm thy

Cch lm

Duyt ton b cc phn t trong mng

Nu a[i] bng (ging) k th lu li ch s i

S dng mt bin xc nh tm thy hay khng tm thy

21


Phn tch

Duyt ton b cc phn t

Vng lp for (while, do while)

Lu li i nu a[i] bng (ging) k

S dng mng lu ch s

Bin xc nh tm thy hay khng tm thy

Bin nhn gi tr 0 hoc 1

Bin nhn gi tr 0 hoc >=1 (tm thy th tng gi tr)

22


#include

#include

void main(){

int a[100], chi_so[100];

int n;//n la s phn t trong mng

int i, k, kiem_tra;

printf( Nhap vao so phan tu cua mang: );

scanf(%d,&n);

printf(Nhap vao gi tr tim kiem);

scanf(%d,&k); 23


kiem_tra = 0;

// Duyt qua tt c cc phn t

for(i = 0;i


if(kiem_tra > 0){

printf(Trong mang co %d phan tu co gia tri bang %d,kiem_tra,k);

printf(\nChi so cua cac phan tula:);

for(i = 0;i < kiem_tra;i++)

printf(%3d,chi_so[i]);

} else

printf(\n Trong mang khong co phan tu nao co gia tri bang %d,k);

getch();}

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9.1.5. Sp xp mng

Bi ton

Cho mng a gm n phn t. Sp xp cc phn t ca mng a theo th t tng dn/gim dn

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9.1.5. Sp xp mng

Gii thut sp xp

Sp xp thm dn (insertion sort)

Sp xp la chn (selection sort)

Sp xp ni bt (bubble sort)

Sp xp vun ng (heap sort)

Sp xp nhanh (quick sort)

Sp xp trn (merge sort)

.

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9.1.5. Sp xp mng

Gii thut sp xp la chn

Tm phn t nh nht cha c sp xp trong mng

i ch n vi phn t u tin trong phn cha c sp

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89.1.5. Sp xp mng

tng

Ln sp xp th 1

So snh a[0] vi cc a[i], i = 1..n-1

a[0] > a[i] => i ch a[0] v a[i]

Thu c a[0] l phn t nh nht

Ln sp xp th 2

So snh a[1] vi cc a[i], i = 2..n-1

a[1] > a[i] => i ch a[1] v a[i]

Thu c a[1] l phn t nh th 2

29

9.1.5. Sp xp mng

tng

Ln sp xp th k

So snh a[k-1] vi cc a[i], i = k..n-1

a[k-1] > a[i] => i ch a[k-1] v a[i]

Thu c a[k-1] l phn t nh th k

..

Ln sp xp th n-1

So snh a[n-2] v a[n-1]

a[n-2] > a[n-1] => i ch a[n-2] v a[n-1]

Thu c a[n-2] l phn t nh th n-1 => cn li a[n-1] l phn t nh th n (ln nht)

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9.1.5. Sp xp mng

A = { 12, 5, 3, 4 };

Lt 1 Lt 2 Lt 3

12 3 3 35 12 4 4

3 5 12 54 4 5 12

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9.1.5. Sp xp mng

//Khai bao cac bienint a[100];int i, j, temp;

//Sap xepfor (i = 0; i < n-1; i++){

for (j = i+1; j a[j]){

tmp = a[i];a[i]= a[j];a[j]= tmp;

} 32

99.1.5. Sp xp mng

V d

Nhp vo t bn phm mt mng s nguyn m trong s phn t cng c nhp t bn phm

Hin th cc phn t va c nhp vo

Sp xp mng m theo th t tng dn trong c hin th cc phn t trong mi lt sp xp.

33

9.1.5. Sp xp mng

#include

#include

void main(){

int m[100];

int n; // n la s phn t trong mng

int i, j, k;

// Nhp gi tr d liu cho mng m

printf( Cho biet so phan tu co trong mang: );

scanf(%d,&n);

34

9.1.5. Sp xp mng

// nhp gi tr cho cc phn t

for(i = 0;i

10

Ni dung

9.1. Mng

9.2. Xu k t

9.2.1. Khi nim xu k t

9.2.2. Khai bo v s dng xu

9.2.3. Cc hm x l k t

9.2.4. Cc hm x l xu

37


Xu k t (string) l mt dy cc k t vit lin tip nhau

di xu l s k t c trong xu

Xu rng l xu khng c k t no

V d: Tin hoc, String

Lu tr: kt thc xu bng k t \0 hay NUL (m ASCII l 0)

38

T i n h o c \0


So snh

Xu k t v mng k t?

Tp hp cc k t vit lin tip nhau

S khc bit: xu k t c k t kt thc xu, mng k t khng c k t kt thc xu

Xu k t A v k t A?

A l 1 k t

A l 1 xu k t, ngoi k t A cn c k t \0 => gm 2 k t

39


a. Khai bo xu

C php

char tn_xu [s_k_t_ti_a+1];

Lu :

lu tr mt xu c n k t chng ta cn mt mng c kch thc n+1

V d

lu tr xu Tin hoc chng ta phi khai bo xu c s phn t ti a t nht l 8

char str [8];40

11


b. Truy cp vo mt phn t ca xu

C php:

tn_xu [ch_s_ca_k_t]

V d

char quequan[10];

Gi s xu ny c ni dung l Ha noi

quequan[0] lu tr H

quequan[1] a

quequan[6] i

quequan[7] \041


Tp tiu s dng: ctype.h

int toupper(int ch): chuyn k t thng thnh k t hoa

toupper(a) => A

int tolower(int ch): chuyn k t hoa thnh k t thng

tolower(B) => b

42


int isalpha(int ch): kim tra xem k t c phi ch ci hay khng (az,A,..Z)

int isdigit(int ch): kim tra ch s (0,1,..9)

int islower(int ch): kim tra ch thng

int isupper(int ch): kim tra ch hoa

int iscntrl(int ch): kim tra k t iu khin (0-31)

int isspace(int ch): kim tra k t du cch (m 32), xung dng (\n 10), u dng (\r 13), tab ngang (\t 9), tab dc (\v 11)

tr v khc 0 nu ng, ngc li tr v 043


#include

#include

#include

void main(){

char ch;

printf(Nhap vao mot ki tu: );

scanf(%c, &ch);

44

12


if(isupper(ch)){

printf(Ki tu nay la chu hoa\n);

printf(Ki tu chu thuong tuong ung %c\n,tolower(ch));

}else if(islower(ch)){

printf(Ki tu nay la chu thuong\n);

printf(Ki tu chu hoa tuong ung %c\n,toupper(ch));

}

getch();

}45


Vo ra xu k t

Tp tiu : stdio.h

Nhp xu k t

gets(tn_xu);

scanf(%s,&tn_xu);

Hin th xu k t

puts(tn_xu);

printf(%s,tn_xu);

S khc nhau gia gets v scanf?

46

9.2.4. Cc hm x l xu k t

Tp tiu : string.h

size_t strlen(char[] tn_xu): tr v di xu

char[] strcpy(char[] xu_ch, char[]xu_ngun): sao chp xu

int strcmp(char[] xu_th_nht, char[]xu_th_hai): so snh hai xu

gi tr 0 : hai xu ging nhau

gi tr0: xu th nht ln hn xu th hai47


char[] strcat(char[] xu_ch, char[]xu_ngun): ghp ni xu ngun vo ngay sau xu ch

Tp tiu : stdlib.h

int atoi(char[] str): chuyn mt xu k t thnh mt s nguyn tng ng

int atol(char[] str): chuyn thnh s long int

float atof(char[] str): chuyn thnh s thc

Khng thnh cng c 3 hm: tr v 0

48

13


#include

#include

#include

void main(){

clrscr();

char str1[10] = abc;

char str2[10] = def;

printf( str1: %s,str1);

printf(\n str2: %s,str2);

printf(\n strcmp(str1,str2)= %d, strcmp(str1,str2)); 49


printf(\n strcpy(str1,str2) = %s, strcpy(str1,str2));



strcpy(str1,ab);strcpy(str2,abc);



printf(\n strcmp(str1,str2) = %d, strcmp(str1,str2));

getch();

}50

Tho lun

51

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