ࡱ> q` ~bjbjqPqP 8 ::9 <<<<<<<PPPP8$P>2$:::: =======$?hcBD=< U | =<<::> <:<:= = 2<<t4: B#P V2= >0>2BQ.B4t4B<t4D U ^ L ==^ > PPP<<<<<< UBND HUYN QU SN PHNG GD&TK THI HC SINH GII LP 9 CP HUYN NM HC 2014-2015 Mn: Tin hc - Lp 9 Thi gian lm bi: 150 pht (Khng k thi gian giao )  CHNH THC - VNG I Bi 1 (4,0 im): Dy s. Vit chng trnh cho php nhp dy s gm n (n> 0) s nguyn: a1,a2,...an. Hy cho bit trong dy s va nhp c bao nhiu s ln hn trung bnh cng ca dy s. Gi a v b ln lt l s ln nht v s b nht ca dy trn. Hy a ra cc s nguyn theo th t tng dn thuc on [b,a] m khng xut hin trong dy s. Nhp s t nhin k. Cho bit trong dy s c k s dng ng cnh nhau khng? Cc s xut hin nhiu ln trong dy l cc s tha cn phi loi b ch gi li mt s. Thc hin in ra dy s va nhp sau khi  loi b cc s tha. V d: Nhpn = 7 3; -2; 4; 3; 2; -3; 7 k =3Xuta) 4 b) -1; 0; 1; 5; 6 c) CO d) 3; -2; 4; 2; -3; 7Gii thcha) TBC =(3+ -2+ 4+ 3+ 2+-3+ 7):7 = 2 nn c 4 s ln hn 2 l: 3; 4; 3; 7 b) T s nh nht (=-3) n s ln nht (=7) cc s trn tha iu kin c) C 3 s dng ng cnh nhau l 4; 3; 2 d) S 3  v tr th 4 lp li nn loi khi dy. Bi 2(3,0 im): Liy tha. Vit chng trnh cho php nhp s nguyn dng N ri thc hin: a) Cho bit N c phi l liy tha 3 ca mt s t nhin khng? b) Vit N di dng mt liy tha vi s mi l s t nhin ca 5. Vit KHONG nu N khng phi l liy tha ca 5. c) Tm s d khi chia NN cho 7. V d: NhpXutGii thchN=8a) CO b) KHONG c) 18 l liy tha 3 ca mt s (8 = 23) 8 khng l liy tha ca 5 88 chia 7 d 1N=25a) KHONG b) 5^2 c) 425 khng l liy tha 3 ca mt s. 25 l mt liy tha ca 5 (25 = 52) 2525 chia 7 d 4 Bi 3 (3,0 im): H trc ta . Vit chng trnh cho php nhp ta  hai im A, B trn mt phng ta  Oxy. Cc im A, B khng nm trn cc trc ta . Kim tra cc im A, B c nm trn cc trc ta  khng? Yu cu nhp li nu im A hoc im B nm trn trc ta . Kim tra cc tnh cht: i xng qua trc tung, i xng trc honh, i xng qua gc ta  ca hai im A, B. Cho bit on thng AB ct h trc ta  (trc tung, trc honh) ti my im? V d:  hnh v bn c: - Hai im B, C i xng qua trc tung. - Hai im A, D i xng qua gc ta . - Hai im A, B cng nm trong mt phn t mt phng. - Hai im C v D khng cng nm trong mt phn t mt phng. - on thng AB ct h trc ti 0 im. - on thng CD ct h trc ti 1 im. - on thng BD ct h trc ti 2 im. - on thng AD ct h trc ti 1 im. ============ HT=========== UBND HUYN QU SN PHNG GD&TK THI HC SINH GII LP 9 CP HUYN NM HC 2014-2015 Mn: Tin hc - Lp 9 Thi gian lm bi: 150 pht (Khng k thi gian giao )  CHNH THC - VNG II Bi 1 (4,0 im): S nguyn t. Vit chng trnh cho php nhp s nguyn dng N ri thc hin. Vi mi s nguyn dng N k hiu S(N) l tng tt c cc s nguyn t khng vt qu N. Hy tnh S(N) vi N c nhp. Hai s nguyn t c gi l cp s nguyn t h hng nu chng hn km nhau 4 n v. Hy in ra cc cp s nguyn t h hng c cc s nguyn t khng vt qu N. V d: (3, 7), (7, 11), (13, 17), (19, 23), (37, 41) ... l cc cp s nguyn t h hng. S nguyn t i xngl mt HYPERLINK "http://vi.wikipedia.org/wiki/S%E1%BB%91_nguy%C3%AAn_t%E1%BB%91" \o "S nguyn t" s nguyn tbng trung bnh cng ca 2 s nguyn t lin trc v lin sau n. In ra cc s nguyn t i xng khng vt qu N. V d: 5 l s nguyn t i xng v 5 = (3 + 7):2 S nguyn t P c gi l s nguyn t o ngc nu vit cc ch s ca P theo chiu ngc li th c mt s mi cing l s nguyn t. In ra cc s nguyn t o ngc khng ln hn N. V d: 13 l s nguyn t o ngc v 13 v 31 u l cc s nguyn t. V d: NhpXutGii thchN= 7a) S(N) = 17 b) NTHH: (3, 7) c) NTX: 5 d) NTN: 2; 3; 5; 7S(7) = 2+3+5+7 = 17. (3, 7) l cp s nguyn t h hng. 5 = (3 + 7): 2. Cc s c 1 ch s th s i xng ca n l chnh n.N=100a) S(N) = 1060 b) NTHH: (3, 7); (7, 11); (13, 17); ... (79, 83) c) NTX: 5; 53 d) NTN: 2; 3; 5; 7; 11; 13; ... 79; 97S(15)=2+3+5+ ... + 97 =1060 (3, 7); (7, 11); (13, 17); ... (79, 83) l cc cp s nguyn t h hng. 5 = (3 + 7): 2; 53 = (47 + 59): 2 11 o ngc cing c 11 ... 79 o ngc c 97 u l s nguyn t. Bi 2 (3,0 im): Ch s. Vit chng trnh cho php nhp mt s nguyn dng N (N<106). Cho bit c th sp xp li cc ch s ca s N  c s chia ht cho 5 khng? (In ra mn hnh CO hoc KHONG). Thc hin xa mt s ch s bn phi ca s N  c s chia ht cho 9. In ra cc s . In ra KHONG nu khng th xa  c s chia ht cho 9. Thc hin sp xp li cc ch s ca s N  c s c gi tr ln nht. V d: NhpXutGii thchN=2014a) CO b) KHONG c) 4210C th sp li 4210; 1240 ... chia ht cho 5. Ln lt xa, cc s 2014; 201; 20; 2 u khng chia ht cho 9. 4210 l s ln nht trong tt c cc s sp xp c.N= 19441a) KHONG b) 1944 c) 94411Khng th sp  c s chia ht cho 5. S 1944 (xa ch s 1 bn phi) chia ht cho 9. 9441 l s ln nht trong tt c cc s sp xp c. Bi 3: (3,0 im) Tam gic Vit chng trnh cho php nhp s nguyn N l chu vi ca mt tam gic c cc cnh u l cc s nguyn. In ra s o ba cnh ca tam gic. Thng bo KHONG nu khng c tam gic c cnh l s nguyn v c chu vi l N. Trong cc tam gic  trn (Cu a) c tam gic vung khng? Trong cc tam gic  trn (Cu a) hy cho bit din tch ca tam gic c din tch ln nht? V d: NhpXutGii thchN= 3a) (1; 1; 1) b) KHONG c) 0.43C 1 tam gic. Tam gic (1; 1; 1) khng vung. 0.43 l din tch ln nht ca tam gic. N= 4a) KHONG b) KHONG c) 0.0Khng c tam gic c cnh nguyn no c chu vi bng 4. Khng c tam gic nn khng c tam gic vung. Khng c tam gic nn xut 0.0.N=7a) (1; 3; 3); (2; 2; 3) b) KHONG c) 1.98C 2 tam gic. Khng c tam gic vung trong 2 tam gic trn. 1.98 l din tch ln nht.N=12a) (2; 5; 5); (3; 4; 5); (4; 4; 4) b) CO c) 6.93C 3 tam gic. C 1 tam gic vung l (3; 4; 5). 6.93 l din tch ln nht. ============ HT=========== UBND HUYN QU SN PHNG GD&TK THI HC SINH GII LP 9 CP HUYN NM HC 2014-2015 Mn: Tin hc - Lp 9 Thi gian lm bi: 150 pht (Khng k thi gian giao ) HNG DN CHM - VNG I Bi 1 (4,0 im): Dy s. a. Code: CuCodeimKhai bo, nhp dyProgram Day_so; Var M:array[1..20] of integer; i,j,n,k, Max, Min:integer; S:real; OK, OOK:Boolean; Begin Repeat Write('Nhap n:');Readln(n); until n>0 For i:=1 to n do Begin Write('M[',i,']=');Readln(M[i]); End;a) m s phn t ln hn trung bnh cng. S:=0; For i:=1 to n do S:=S+M[i]; S:=S/n; k:=0; For i:=1 to n do if M[i]>S then k:=k+1; Writeln('a) So phan tu lon hon TBC la: ',k);b) Cc s nguyn trong on [ Min; Max] khng c mt trong dy Write('b) Cac so can tim: '); Max:=M[1];Min:=M[1]; OOK:=false; For i:=1 to n do Begin if M[i]>Max then Max:=M[i]; if M[i]<Min then Min:=M[i]; End; For k:=Min to Max do Begin OK:=true; For i:=1 to n do if k=M[i] then OK:=false; if OK then Begin Write(k,'; '); OOK:=true; end; End; If OOK then Writeln else Writeln('KHONG');&(HLrt H L N P ~    Ϲğyi^V^K^hh fmHsHhC(mHsHhhAkmHsHhAkhAk56aJmHsHh256aJmHsHhF(56mHsHhah.5CJ\aJh<h.5CJ\aJh9Ih.6CJaJh9Ih.CJaJh.5CJ\aJh9Ih.5CJ\aJ.jh9Ih.5CJU\aJmHnHuhX)h.CJaJ&@J L N P ~ N $`a$gdAk$a$gdAk$a$gd.Skd$$Ifl0 & 7 tI)644 la7 $$Ifa$gd. b|  4 6 : < F H N j H N 4 8 ` b ",.Z\frxz&ûí{{{{shjmHsH hhC hhpXohhCmHsHhhpXomHsHhzDFhAk56mHsHhzDFhpXo56mHsHh`mHsHhh wmHsHhfmHsHhC(mHsHhhAkH*mHsHhhAkmHsHhhmHsH)N < ".\fhY|kd$$Ifl0$M! t0644 la $IfgdpXo$a$gdAk$ & F h^`ha$gd|z hr|jy|kd$$Ifl0$M! t0644 la $IfgdpXo &*02hnxHf6J|(46jyng` hhF P hhENxhhENxmHsHhhENx56mHsHhWH*mHsHhbsEmHsHh$GmHsHhWmHsHhhAkmHsHhh256mHsHhhAk56mHsHhJmHsHhhpXomHsHhhCmHsHhjmHsHhhjmHsH%(6@J`zznnnni``` $IfgdX5gdX5 $`a$gdAk$a$gdAk|kd]$$Ifl0$M! t0644 la `bjx.qhhhhhhh $IfgdX5kd$$IflF$v  t06    44 la:LXZbd6VXZ\VXHZbtxz|*|u|q|q|h- h-h- h-hX5hXhdq hdqhdq hdqhX5 hhX5hh256hhX556 hD&56 h!RUH* hh@h!RUh!RUH*mHsHhhENxmHsHhh@mHsH hhF PhhENxH* hhENx+.0:LZdqhhhhhhh $IfgdX5kd5$$IflF$v  t06    44 laZXH*NqlllaaalXX $IfgdX5 $ & Fa$gd)_gdX5kd$$IflF$v  t06    44 la *^d|xz| |qf[fPfPhhF(mHsHhhCmHsHhh.mHsHhhi=mHsHjhhVUmHsHhh.:mHsHh$w]mHsHh.:mHsHhVmHsHhc[mHsHhhVmHsHh-hX556mHsHh-hV56mHsH h -hX5 h<:BhCW h<:BhA2 h<:Bh<:B h<:BhX5N (z~yqq$a$gd7gdX5xkd $$Ifl0$ t0644 la $IfgdX5  "$&(*Pjt $$Ifa$gd.$a$gd7$a$gdC "*PR:rvxzõמwiw^PB4hh'.56mH sH hhi=56mH sH hh256mH sH h{thi=mHsHh{thC5\mHsHh{th.5\mHsHhh5\mHsHh.5\mHsHhh.6mHsHh ]5\mHsHhh.5\mHsH&jhh.5U\mHnHuhh.mHsHhhVmHsHhhLfmHsHh8BmHsHtvxzp` \!idgdKU$ & F h^`ha$gd7Z$ & F h^`ha$gd ]$a$gd7$a$gd.Skd $$Ifl0 & 7 tI)644 la7 4Lpt6` \ \!!!T"V"n"p"""8#Z#\###$ûûûè~~~vnf[hhEimH sH h'.mH sH h]mH sH h"mH sH jh0Ah0AUmH sH h0Ah0AmH sH h*H t0%6    44 la $IfgdjJ$a$gdC 4555$5&585:5H556`WWWWWWWW $IfgdjJkd$$$IflFD%>*H t0%6    44 la $If^` gdjJ 6T6V6^666666.7f7hkd%$$IflFD%>*H t0%6    44 la $IfgdjJ f7h7r7777768n8qhhhhhhh $IfgdjJkd%$$IflFD%>*H t0%6    44 lan8p8r8t88888888qiaYWWWWWW$a$gdC$a$gdAk$a$gdCkd&$$IflFD%>*H t0%6    44 la 888888h99: : :&:::<:n:p::,;j;;;;;ƴޤxk[TITB;B h bh b h bhTLhQ^hTLmHsH h~hTLhAkhTL56aJmHsHhTL56aJmHsHhTL56mHsHhTL5CJ\aJmHsH"hTLhTL5CJ\aJmHsHhTLhTL6CJaJmHsH"hTLhTL5CJ\aJmHsH.jh9IhTL5CJU\aJmHnHuhTLhTLCJaJmHsHhh*mHsHhTLmHsH8888D9h99::: :::<:p:::$If$a$gdTL$a$gdTLSkdx&$$Ifl0 & 7 tI)644 la7 $$Ifa$gdZ::::::$;b;z;;_VVVVV $IfgdTLkd'$$IflFpX $8 t0$6    44 la $$Ifa$gdr($If ;;<<<P<<<<<=\S $Ifgdr(kd'$$IflFpX $8 t0$6    44 la $$Ifa$gdr( $IfgdTL ;<<<<<< ====N>T>>>?F?^?h??6ABA^A~AAAAAAAfh؜ܜΞҞPFH~븭䠙{hehhe56mHsHh/oh9h2K h2KhTL h2Kh2KUh bh bh bmHsHh bhTLmHsHhQ^hTLmHsHhQ^hTLmHsHhTL hTLhr( hTLhTL h~hTL h bhTL h bh b.="=b=z====P>R>T>\kd'$$IflFpX $8 t0$6    44 la $$Ifa$gdr( $IfgdTL T>>?`????.@@@r@@@AAAAA $$Ifa$gdr( $IfgdTL $Ifgdr(AAh0fz Hqh________ $IfgdTL $Ifgdr(kdz($$IflFpX $8 t0$6    44 la c) Kim tra tnh cht c k s dng ng cnh nhau. Write('Nhap so k: ');Readln(k); Write('c) Co k so duong dung canh nhau: '); OOK:=false; For i:=1 to n - k+1 do Begin OK:=true; For j:=i to i+k-1 do if M[j]<0 then OK:=false; if OK then OOK:=true; End; if OOK then Write('CO') else Write('KHONG'); Writeln; d) Loi cc phn t tha ca dy Write('d) Sau khi loai PT thua: '); For i:=1 to n do Begin ok:=true; For j:=1 to i-1 do if M[i] = M[j] then OK:=false; if OK then write(M[i],'; '); End; Readln; End.b) Yu cu xy dng Test: Cn xy dng t nht 4 test  chm. Mi test ng cho 0,25 im/cu Trong  ch xy dng mt s Test c bit: n = 0 Phn t Max = Min. C phn t ln ln hn trung bnh cng. Khng c phn t ln hn trung bnh cng. C k phn t dng cnh nhau. Khng c k phn t dng cnh nhau. Bi 2(3,0 im): Liy tha. Khai bo, nhp NProgram Luy_thua; Var N,P,i,r:integer; Begin Write('Nhap so N:'); Readln(N);a) Kim tra N c l lp phng ca mt s nguyn.Write('a) Xet N la luy thua 3:'); P:=1; While P*P*P<N do P:=P+1; if P*P*P = N then Writeln('CO') else Writeln('KHONG');b) Kim tra N c l mt liy tha ca 5 khng? Vit N dng mt liy tha ca 5.P:=1; i:=0; While P<N do Begin P:=P*5; i:=i+1; end; Write('b) So N la luy thua cua 5:'); if P = N then Writeln('N= 5^',i) else Writeln('KHONG');c) Tm s d ca NN chia 7.r:=N mod 7; For i:=1 to N-1 do Begin r:=(r*N) mod 7; End; Writeln('c) So du la: ',r); Readln; End. Bi 3 (3,0 im): H trc ta . a) Code: a) Khai bo, nhp, kim tra ta  hai imProgram He_Toa_Do; Var Xa,Ya,Xb,Yb:real; d:integer; Begin repeat Writeln('Nhap toa do diem A:'); Write('H.Do:');Readln(Xa); Write('T.Do:');Readln(Ya); until (Xa<>0) and (Ya<>0); Repeat Writeln('Nhap toa do diem B:'); Write('H.Do:');Readln(Xb); Write('T.Do:');Readln(Yb); until (Xb<>0) and (Yb<>0);b) Kim tra tnh cht i xng trc qua cc trc ta ; i xng tm qua gc ta .Writeln('a) Xet tinh chat doi xung: '); Write('Doi xung truc tung:'); if (Ya=Yb) and (Xa=-Xb) then Writeln('CO') else Writeln('KHONG'); Write('Doi xung truc hoanh:'); if (Ya=-Yb) and (Xa=Xb) then Writeln('CO') else Writeln('KHONG'); Write('Doi xung goc toa do:'); if (Ya=-Yb) and (Xa=-Xb) then Writeln('CO') else Writeln('KHONG');c) Tm giao im ca on thng AB vi cc trc ta .d:=0; if Xa*Xb<0 then d:=d+1; if Ya*Yb<0 then d:=d+1; if (d=2) and (Ya/Yb = Xa/Xb) then d:=1; Writeln('c) Doan thang AB cat he truc tai ',d, ' diem'); readln; End.b. Xy dng test: Cn xy dng t nht 4 test  chm. Mi test ng cho 0,25 im/cu Khi xy dng test cn  cc trng hp sau: C/Khng i xng trc tung. C/Khng i xng trc honh. C/Khng i xng gc ta . C/Khng cng nm trong mt gc phn t. S giao im AB vi trc ta  l 0; 1; 2. Gc ta  thuc on thng AB. UBND HUYN QU SN PHNG GD&TK THI HC SINH GII LP 9 CP HUYN NM HC 2014-2015 Mn: Tin hc - Lp 9 HNG DN CHM - VNG II Bi 1 (4,0 im): S nguyn t. Khai bo binProgram Nguyen_to; Var N,a,b,i:integer; P:longint; ok:boolean;Hm kim tra s nguyn t.Function NT(N:integer):Boolean; Var i:integer; Begin NT:=n>1; For i:=2 to n-1 do if n mod i = 0 then NT:=false; End;Hm tm s o ngc (cc ch s)Function DAO(N:integer):integer; Var tam:integer; Begin Tam:=0; While N>0 do Begin Tam:=Tam*10+(N mod 10); N:=N div 10; End; DAO:=Tam; End;a) Tng cc s nguyn t khng qu NBegin Write('Nhap so N: ');readln(N); P:=0; For i:=2 to N do if NT(i) then P:=P+i; Writeln('a) S(N)= ',P);b) Cc cp s nguyn t h hng. Write('b) So nguyen to ho hang: '); ok:=false; For i:=2 to N-4 do if NT(i) and NT(i+4) then Begin Write('(',i,';',i+4,');'); ok:=true; end; if not ok then writeln('KHONG CO') else Writeln;c) S nguyn t i xng. ok:=false; Write('c) So nguyen to doi xung: '); For i:=5 to N do Begin a:=i-1; While not NT(a) and (a>0) do a:=a-1; b:=i+1; While not NT(b) do b:=b+1; if NT(i) and NT((a+b) div 2) then Begin Write(i,';'); ok:=true end; end; if not ok then Writeln('KHONG CO') else Writeln;d) Tm cc s nguyn t o ngc. ok:=false; Write('d) So nguyen to dao nguoc: '); For i:=2 to N do if NT(i) and NT(DAO(i)) then Begin Write(i,'; '); ok:=true end; if not ok then writeln('KHONG CO') else writeln; Readln; End. Cn xy dng t nht 4 test  chm. Mi test ng cho 0,25 im/cu Bi 2 (3,0 im): Ch s. Yu cuCodeimKhai bo bin, nhp NProgram chu_so; Var N, NN, S,i,j,k, tam: integer; OK:boolean; M: array[1..10] of byte; Begin Repeat Write('Nhap so N:');Readln(N); until n>=0;a) Xt N c th sp xp li  c s chia ht cho 5 khng?NN:=N; OK:=false; While NN>0 do Begin if NN mod 5 =0 then ok:=true; NN:=NN div 10 End; Write('a) Chia het cho 5:'); If OK then Writeln('CO') else Writeln('KHONG');b) Xt N c th xa mt s ch s bn phi  c s chia ht cho 9 khng.OK:=false; NN:=N; While NN>0 do Begin if NN mod 9 = 0 then Begin OK:=true; Write(NN,'; '); NN:=NN div 10; End; Write('b) Chia het cho 9:'); If OK then Writeln else Writeln('KHONG');c) Sp xp li cc ch s ca N  c s c gi tr ln nht.NN:=N;k:=0; While NN>0 do Begin k:=k+1; M[k]:=NN mod 10; NN:=NN div 10; End; For i:=1 to k do For j:=i+1 to k do if M[i]<M[j] then Begin Tam:=M[i]; M[i]:=M[j]; M[j]:=Tam; End; Write('c) So lon nhat xep duoc la:'); For i:=1 to k do Write(M[i]); Readln; End. b) Yu cu xy dng Test: Cn xy dng t nht 4 test  chm. Mi test ng cho 0,25 im/cu Trong  cn kim tra c cc trng hp: C/Khng th sp xp li  c s chia ht cho 5. C/Khng th xa cc ch s bn phi  c s chia ht cho 9. Bi 3: (3,0 im) Tam gic a. Code: Khai bo bin, nhp N.Program Tam_giac; Var N,a,b,c:integer; p,S,H,SM:real; OK:boolean; Begin Write('Nhap so N:');Readln(N);a) Cc tam gic cnh l s nguyn c chu vi N. Write('a) Cac tam giac co the la:'); For a:=1 to n-2 do For b:=a to n - a do For c:=b to n - a-b do if (c<(a+b)) and (a+b+c=N) then write('(',a,';',b,';',c,'); '); Writeln;b) Xt: C tam gic vung cnh l s nguyn chu vi N khng. Write('b) Tam giac vuong chu vi N:'); OK:=false; For a:=1 to n-2 do For b:=a to n - a do For c:=b to n - a-b do if (c<(a+b)) and (a+b+c=N) and(a*a + b*b = c*c) then OK:=true; if OK then Writeln('CO') else Writeln('KHONG');c) Tm din tch tam gic ln nht SM:=0; For a:=1 to n-2 do For b:=a to n - a do For c:=b to n - a-b do if (c<(a+b)) and (a+b+c=N) then Begin p:=(a+b+c)/2; S:=sqrt(p*(p-a)*(p-b)*(p-c)); if S>SM then SM:=S; End; Writeln('c) Dien tich lon nhat la: ',SM:2:2); Readln; End.b. Xy dng test: Cn xy dng t nht 4 test  chm. Mi test ng cho 0,25 im/cu Cc test cn lu cc trng hp: C/Khng c tam gic cnh nguyn c chu vi N (V d: N = 1; 2; 4). C/Khng c tam gic vung.     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