1/ Vit hm2/ Vit chng trnh

cu lnh iu kin --> cu lnh lp --> mng --> chui --> cu trc

Bi 1: Kin thc v vng lpn = 3=> S = 1 + 1 * 2 + 1 * 2 * 3

n = 4=> S = 1 + 1 * 2 + 1 * 2 * 3 + 1 * 2 * 3 * 4

S = tch t 1 ti 1 + tch t 1 ti 2 + tch t 1 ti 3 + tch t 1 ti 4

n = 6=> S = 1 + 1 * 2 + 1 * 2 * 3 + 1 * 2 * 3 * 4 + 1 * 2 * 3 * 4 * 5 + 1 * 2 * 3 * 4 * 5 * 6

= 1 + 2 + 6 + 24 + 120 + 720 = 873

Bi 2: Kin thc v chuiabcdcba => i xngabcdxya => khng i xng

Th no mi gi l i xng ?

c chui t tri qua phi hoc t phi qua tri th u nhn 1 kt qu nh nhau.

Ch chui ra lm i ri ly tng phn t bn tri so khp bn phi, nu tt c u ging nh

0 1 2 3 4 5 6a b c d c b b => di l 7 (s l)

so snh tng cp nh sau:0 vi 61 vi 52 vi 4

0 1 2 3 4 5 6 7a b c d d c b a => di l 8 (s chn)

so snh tng cp nh sau:0 vi 71 vi 62 vi 53 vi 4

iu th 1 ta thy: v tri + v phi = di - 1 => v phi = di - 1 - v triiu th 2 ta thy: Bn v tri lun bt u chy t 0 v kt thc l : ( di / 2) - 1

Bi 3: Kin thc v mng 1 chiu.

* XA *

u tin ta c mng a c 7 phn t gm: 1, 2, 3, 4, 5, 6, 7

tc l nh sau:

=> ch s (v tr): 0 1 2 3 4 5 6=> phn t: 1 2 3 4 5 6 7

ta xa i v tr 2 tc l s 3

=> mng cn li sau khi xa l:

=> ch s: 0 1 2 3 4 5=> phn t: 1 2 4 5 6 7

B1: Xc nh mi lin h gia mng trc khi xa v mng sau khi xaB2: Tm cng thc (xc nh chiu chy ca n)

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B1:Sau | Trca[2] | a[3]a[3] | a[4]a[4] | a[5]a[5] | a[6]

B2:C 2 chiu chy l t trn xung di hoc t di ln trn. Ta chn chiu m n khng b trn

=> Ta thy nu chn chiu chy t di ln trn th s b trng (c th l a[5], a[4], a[3], ata khng chn chiu ny m chn chiu ngc li tc l t trn xung di.

Lc ny ta chn theo tay tri. Tc l chy vng lp t 2 cho ti 5m ta khng c code cng 2 vi 5 m phi to ra cng thc cho n.

ta thy 2 chnh l v tr xa ban uta thy 5 chnh l n - 2

=> ta c vng lp sau:

for(int i = ViTriXoa; i < n - 1; i++){

a[i] = a[i + 1];}

0 1 2 3 4 5-1 -2 -3 4 5 -6

xa v tr 0=> mng cn li l: 0 1 2 3 4-2 -3 4 5 -6

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