tut1 solution

8/11/2019 Tut1 Solution

1/14

Khoa Khoa hc v Kthut My tnhBmn Khoa hc My tnh

1

KTHUT LP TRNHBi tp Tun 2

Prototype v Flow Chart

Hng dn:

Prototype ca mt vn sc biu din nh sau:

outfname(in1, in2,..., inn)

Trong outl dliu xut, fname l tn ca vn , inil cc dliu nhp ca bi ton.

V d:

Vn :Tnh tng cca hai snguyn av b

Bi ton ny c dliu nhp l av b, dliu xut l c. Nh vy prototype ca bi tonsl nh sau:

ctong(a,b)

Mrng (Cha yu cu phi lm trong phn bi tp ny):

Trong cc bi hc sau, khi hc vcch khai bo kiu d liu ca bin, tn ca d liunhp xut sc thay bng kiu, cc dliu nhp cng sc khai bo vi kiu tngng. Lc prototype sl

inttong(inta, intb)

Bi tp bt buc

Xy dng prototype v vflow chart cho cc bi tp sau


2/14


2

Bi 1.Tnh tng d ca ba snguyn a,bv cPrototype: d tong(a,b,c)Flowchart:

Hnh 1. Tnh tng 3 snguyn a,b,c

Start

Input a,b,c

da+b+c

Print d

Stop


3/14


3

Bi 2. Tnh chu viPca mt hnh trn c bn knhRPrototype:P chuvi(R)Flowchart:

Hnh 2. Tnh chu vi hnh trn bn knh R

Start

Input R

P

2*3.14*R

Print P

Stop


4/14


4

Bi 3.So snh din tch A1 ca mt hnh vung c cnh E vi din tch ca mt hnhtrn c bn knhR(dliu xut ca bi ny sl mt bin B c gi trng hoc sai)

Prototype:B sosanhdientich(E,R)Flowchart: B c gi trng (true) nu A1 A2 (din tch hnh trn)

B c gi trsai (false) nu A1 < A2

Hnh 3. So snh din tch hnh vung cnh E v din tch hnh trn cnh R

Start

Input E,R

A1E*E

A23.14*R*R

Print B

Stop

A1 A2?

BTrue

Yes

No

BFalse


5/14


5

Bi 4.So snh tui ca hai ngi c ngy thng nm sinh ln lt lD1,M1,Y1(i vingi thnht) vD2,M2, Y2(i vi ngi thhai)

Prototype:B sosanhtuoi(D1,M1,Y1,D2,M2,Y2)Flowchart:

Hnh 4. So snh tui

Start

Input

D1,M1,Y1,D2,M2,Y

Stop

diffm0?

diffmM1-M2+(Y1-Y2)*12

diffdD1-D2

diffm>0?

B1

Yes

No

Yesdiffd


6/14


6

Bi tp nng caoBi 5.M tlm thno tnh giai tha ca mt snguyn dng N, N l gi trnhp vo.

Prototype : f giaithua(N)Flowchart :

Hnh 5. Tnh giai tha

Start

Input N

f1

i2

Print f

Stop

i N?

ff*i

ii+1

YesNo


7/14


7

Bi 6.M tgii thut Euclid tm c chung ln nht ca hai snguyn dng, sau , tmbi chung nhnht ca hai snguyn dng.

Tm c schung ln nhtPrototype : uscln(a, b)

Flowchart 1: a,b > 0.

Hnh 6. Tm c schung ln nht ca 2 snguyn dng

Start

Input a,b

ra mod b

ab

br

Print a

Stop

b > 0?Yes

No


8/14


8

Tm bi schung nhnhtPrototype : p bscnn(a, b) ;

Flowchart : b)uscln(a,

|b*a|

b)bscnn(a,

Hnh 7. Tm bi schung nhnht ca 2 snguyn dng

Start

Input a,b

ra mod b

ab

br

Print p

Stop

b > 0?Yes

No

pa*b

pp/a


9/14


9

Bi 7.M tqu trnh tm kim gi trln nht v gi trnhnht trong mt mng c N snguyn. In cc gi trtm c ra mn hnh.

Prototype : minmax(N,a) ;Flowchart :

Hnh 8. Tm min, max trong mng N snguyn

Start

Input N,a

i1

mina

maxa

Print

Stop

i < N?

Yes

ii+1

Input a

a>max?

maxa

Yes No

a


10/14


10

Bi 8.Vit chng trnh nhn vo l mt ng trn tm O(x0, y0) v bn knh R. Kim tra

xem mt im A(x, y) l nm trn, trong hay ngoi ng trn .

Prototype : diemduongtron(x0,y0,R,Ax,Ay) ;

Flowchart :

Hnh 9. Xc nh vtr tng i ca im v ng trn

Start

Input

distsqrt(sqr(Ax-x0)+sqr(Ay-y0))

Print Trong

Stop

dist


11/14


11

Bi 9.Vit chng trnh tnh ng knh, chu vi, din tch to bi ca ng trn bn knh R.

Prototype : duongtron(R) ;Flowchart :

Hnh 10. Tnh ng knh, chu vi, din tch hnh trn bn knh R

Start

Input R

d

2*Rp2*3.14*R

sR*R

Print d,p,s

Stop


12/14


12

Bi 10.Vit chng trnh chuyn i gia mt v inch.Hng dn: 1 inch = 2,54 cm

Prototype : convert(a, sel) ; //sel=1: inch sang mt, sel 1: mt sang inchFlowchart :

Hnh 11. i n vmt v inch

Start

Input sel

ma*2.54/100

Print Chieu dai met: m

Stop

sel=1?Yes

Input a

incha*100/2.54

Input a

Print Chieu dai inch: icnh

No


13/14


13

Bi 11.Vit chng trnh nhn vo nm hp l, kim tra xem c phi l nm nhun haykhng, ghi kt qura mn hnh.

Hng dn: mt nm l nhun khi m n chia ht cho 4 nhng khng chia ht cho 100,hoc l n chia ht cho 400.

Prototype : kiemtranhuan(year)Flowchart :

Hnh 12. Kim tra mt nm l nhun.

Start

Input year

mod4year%4

mod100year%100

mod400year%400

Print Nhun

Stop

(mod4==0&&mod1000)|

|(mod400==0)?

Print Khng nhun

Yes No


14/14


14

Bi 12.Cho di 3 cnh a, b, c, kim tra xem n c phi l 3 cnh ca mt tam gic cn, tamgic u, tam gic vung hay tam gic vung cn khng.

Prototype : tamgiac(a,b,c) ;Flowchart :

Hnh 13. Xc nh c im tam gic to thnh t3 cnh a,b,c

Start

Input a,b,c

Print

Khng l

Stop

((a>|b-c|)&&(a|a-c|)&&(b|b-a|)&&(c

tut1 solution

Documents