ࡱ> []Zbq` 0njbjbjqPqP 8::7////t0=11111444<<<<<<<$>hA^<43444<11x=888411<84<88810 kω/4~8m9d=0=8zA#6zA8zA8\44844444<<8 444=4444*// UBND HUYN QU SN PHNG GD&TK THI HC SINH GII LP 9 CP HUYN NM HC 2010 - 2011 Mn: Tin hc Thi gian lm bi: 150 pht (Khng k thi gian giao )  CHNH THC - VNG I Bi 1: ( 6.0 im) Vit chng trnh cho php nhp mt dy gm n s nguyn t bn phm v thc hin: a. Sp xp v in dy s va nhp theo th t gim dn. b. In ra phn t c gi tr tuyt i ln nht trong dy. c. In ra cc s chnh phng c trong dy. d. Mt phn t ca dy c xem l phn t chnh nu n xut hin [ EMBED Equation.3 ] ln tr ln. Hy kim tra v in ra phn t chnh nu c. V d: Vi n = 6 cho php nhp mt dy gm 6 s. Nu nhp dy s: -5; -2; 0; -2; 4; -2 a. Cho kt qu l: 4; 0; -2; -2; -2; -5 b. Phan tu can tim: -5. c. Phan tu chinh phuong: 0; 4 d. Phan tu chinhl: -2 (V -2 xut hin 3 ln trong dy c 6 phn t) Bi 2: ( 4.0 im) Vit chng trnh cho php nhp t bn phm mt s nguyn dng n v thc hin: a. In ra cc s nguyn t b hn hoc bng n. V d: n = 7 th in 2; 3; 5; 7. b. In ra s nguyn t nh nht khng b hn n. V d: n = 7 th in ra 7. n = 10 th in ra 11. n = 24 th in ra 29. c. In ra cc cp s nguyn t l hai s nguyn l lin tip nh hn hoc bng n. V d: Nhp n = 20 th in ra: 3, 5 5, 7 11,13 17,19 Yu cu lu kt qu lm bi: - Mi cu thc hin lu trong mt file vi tn sau: 1a; 1b; 1c; 1d, 2a; 2b hoc lu nhiu cu trong cng mt file: V d: Cu 1a v 1b lu chung th t tn l 1ab. - Cc file chng trnh c lu  th mc: D:\HSG_TIN\<SDD>\ *.pas. - Ni dung cc file chng trnh c gim th in v th sinh hon thnh phn ghi thng tin th sinh.  UBND HUYN QU SN PHNG GD&TK THI HC SINH GII LP 9 CP HUYN NM HC 2010-2011 Mn: Tin hc Thi gian lm bi: 150 pht (Khng k thi gian giao )  CHNH THC - VNG II Bi 1: (6.0 im) Vit chng trnh cho php nhp t bn phm mt s nguyn dng a v thc hin: a. Cho bit a l s c my ch s. b. Cho bit tng cc ch s ca a. c. Ta gi s  i xng vi a l s nguyn dng thu c t a bng cch o ngc th t cc ch s ca a. Vit chng trnh in ra s i xng ca s nguyn dng a. d. Cho a v b l 2 s nguyn dng. Ta gi tng i xng ca a v b l s i xng vi tng ca s i xng vi a v s i xng vi b. Vit chng trnh cho php nhp hai s nguyn dng a, b v in ra tng i xng ca chng. V d: Nhp a = 25 th cho kt qu l: a. So a co 2 chu so. b. Tong cac chu so cua a la 7. c. So doi xung cua a la 52. d. Nu a = 25 v b = 26 th tng ca s i xng vi a v s i xng vi b l 52 + 62 = 114. V th tng i xng ca a v b l 411. Bi 2:(4.0 im) Tam gic Pascal bc n c l tam gic s c n + 1 dng v c dng: n=0 : 1 n=1 : 1 1 n=2 : 1 2 1 n=3 : 1 3 3 1 n=4 : 1 4 6 4 1 Trong tam gic s ny, gi tr cc s trong hng th n tun theo qui lut sau: - S  ct th n + 1 v ct th nht lun bng 1. - Mi s  ct n n ct 2 bng tng hai s ng  hng trn cng ct v ct trc n. Gi tr cc s  hng n l h s khi khai trin biu thc (x + y)n. Vit chng trnh cho php nhp t bn phm s nguyn n v thc hin: a. In ra tam gic Pascal bc n. b. Khai trin (x + y)n . V d: n = 3 th in: x^3 + 3x^2y^1 + 3x^1y^2 + y^3 n = 5 th in: x^5 + 5x^4y^1 + 10x^3y^2 + 10x^2y^3+ 5x^4y^1 + y^5. Yu cu lu kt qu lm bi: - Mi cu thc hin lu trong mt file vi tn sau: 1a; 1b; 1c; 1d, 2a; 2b hoc lu nhiu cu trong cng mt file: V d: Cu 1a v 1b lu chung th t tn l 1ab. - Cc file chng trnh c lu  th mc: D:\HSG_TIN\<SDD>\ *.pas. - Ni dung cc file chng trnh c gim th in v th sinh hon thnh phn ghi thng tin th sinh.  UBND HUYN QU SN PHNG GD&TK THI HC SINH GII LP 9 CP HUYN NM HC 2010 - 2011 Mn: Tin hc Thi gian lm bi: 150 pht (Khng k thi gian giao ) HNG DN CHM - VNG I Bi 1: (6.0 im) Khai boProgram Cau1abcd; Var M: array[1..20] of integer; n,i,j,tam:integer; xh,chinh:integer; ok: boolean;0.5Nhp dy sBegin Write('Nhap n: &(@  < > B D r t ļsjbZRhCJaJhN|CJaJh'7CJaJh]5CJaJhUh]CJ\aJhBCJaJh8&&h]CJaJh8&&h]6CJaJhhyh]CJaJhUCJaJhSh]CJaJh]CJaJhx;h]5CJaJhh]CJ\aJ+jhhCJU\aJmHnHuhh]5CJ\aJ &@> @ B D r t @ " x $a$gd'7$a$gd]Skd$$Ifl0 %  td&644 la $$Ifa$gdN|nj & ( * , . 6  F P v x z ~ گϐ~lZH~Z~Z"h4h5CJ\aJmH sH "h4h55CJ\aJmH sH "h4hW75CJ\aJmH sH "h4hT5CJ\aJmH sH hW75CJ\aJ'jhh45CJEHU\aJ jlP h45CJUVaJjh5CJU\aJh55CJ\aJh5CJ\aJh'75CJ\aJh'7CJaJhN|CJaJx  P f` $`a$gdTgd?Ay$a$gdW7$a$gd'7$a$gd5 Jb³Ѥn_nPnE:hU5CJ\aJh?Ay5CJ\aJh?Ay5CJ\aJmH sH h O15CJ\aJmH sH "h?Ayh?Ay5CJ\aJmH sH hCJaJmH sH hN|CJaJmH sH h?AyCJaJmH sH h?Ayh?AyCJaJmH sH h,5CJ\aJmH sH h55CJ\aJmH sH "h5h55CJ\aJmH sH hW75CJ\aJmH sH h5h55CJ\aJ.2 &p$ԱscWH8Hh]h]6CJaJmH sH h]h]CJaJmH sH h4CJaJmH sH h]h]5CJaJmH sH h4h]CJ\aJ+jh4h4CJU\aJmHnHuh4h]5CJ\aJhUhB5CJ\aJ(jhB5CJU\aJmHnHuhUh]5CJ\aJhU5CJ\aJhUhU5CJ\aJhUCJaJhUhUCJaJ2 &p "~Skd$$Ifl0 % P td&644 la $$Ifa$gdN|$a$gdU$a$gd'7 $`a$gdv gd?Ay "$TVz` 6tbrgd+H!gdgd39 $1$a$gdCTgdCT$a$gdv gd#-w`gd4$a$gd$a$gd]$TV`bfnxz$J`pz|~¶˜{mbmbmbWmWbmbLWh#-w5CJ\aJhCT5CJ\aJhX5CJ\aJhh#-w5CJ\aJh#-w5CJ\aJ"h4h#-w5CJ\aJmH sH "h4h5CJ\aJmH sH hCJaJmH sH hN|CJaJmH sH h]hCJaJmH sH h]h?AyCJaJmH sH h5h'7CJaJmH sH hUh]CJ\aJmH sH ~Vd $2õxfWfLAh#-w5CJ\aJhCT5CJ\aJhCT5CJ\aJmH sH "h39h#-w5CJ\aJmH sH h395CJ\aJmH sH "h39h5CJ\aJmH sH hv h#-w5CJ\aJhv hv 5CJ\aJhv h395CJ\aJhv 5CJ\aJhCT5CJ\aJhCThCT5CJ\aJhCTh395CJ\aJh45CJ\aJ24npt (6@DJNTd ɾtithZ5CJ\aJh|5CJ\aJhAwCJaJhN|CJaJh!!."0"j"#J$%%B%\%% $$Ifa$gd'gd39$a$gdU`gdUgdUgdXgd~=gd $`a$gdB6J $`a$gdAwgd+H! !!!6!8!>!j!!!!!!!!"$"."0"j"##J$L$%ʸʥʖrrrcTBTBTB"hUhU5CJ\aJmH sH hUhUCJaJmH sH hU5CJ\aJmH sH "h|h+H!5CJ\aJmH sH "h+H!h+H!5CJ\aJmH sH hX5CJ\aJmH sH %hv h+H!5CJH*\aJmH sH "hv h~=5CJ\aJmH sH "hv h+H!5CJ\aJmH sH "h~=h+H!5CJ\aJmHsH"h~=h~=5CJ\aJmHsH%%%B%D%\%% &X&`&|&&&&&&(DZri]TE6E+hhCJaJhhCJaJmH sH hhCJaJmH sH h5CJaJhUhCJ\aJhCJ\aJhh6CJaJmH sH hhCJaJmH sH hh5CJaJmH sH hZhCJ\aJmH sH +jhZhZCJU\aJmHnHu"hZh5CJ\aJmH sH "h+H!hix5CJ\aJmH sH (jh'B5CJU\aJmHnHu%%%Z&\&^&`&&&&&&.'\'' $Ifgd $Ifgd5Pgd39$a$gdSkd?$$Ifl0 %  td&644 la $$Ifa$gd'''''''LHL\LL\S $Ifgd5Pkd$$IflF(# t0 $6    44 la $$Ifa$gd5P $Ifgd ');Readln(n); For i:= 1 to n do Begin Write('M[',i,']='); Readln(M[i]); End;0.5Sp xp dy s For i:=1 to n-1 do For j:=i+1 to n do if M[j]> M[i] then Begin Tam:=M[i]; M[i]:=M[j]; M[j]:=Tam; End;1.5In dy s  sp xp Write('Day so sau khi sap xep: '); For i:=1 to n do Write(M[i],';');0.5Phn t c gi tr tuyt i ln nht Writeln; Write('Phan tu co gia tri tuyet doi lon nhat:'); if abs(M[1])>abs(M[n]) then write(M[1]) else write(M[n]);0.5In cc phn t l s chnh phng Writeln; Write('Cac phan tu la so chinh phuong: '); For i:=1 to n do if M[i]>=0 then if sqr(trunc(sqrt(M[i]))) = M[i] then write(M[i],';');1.0Tm v in ra phn t chnh ca dy. Writeln; ok:=false; For i:= 1 to n do Begin xh:=0; For j:=1 to n do if M[i]=M[j] then xh:=xh+1; if xh >= n div 2 then Begin chinh:=M[i]; ok:=true end; End; if ok then write('Phan tu chinh la: ',chinh) else Write('Khong co phan tu chinh'); Readln; End.1.5 Bi 2: (4.0 im) Khai boProgram Cau2abc; Var n,i: integer; tim: boolean;0.5Hm xem mt s l s nguyn t.Function NT(n:integer):boolean; Var ok: boolean; i: integer; Begin ok:=true; For i:=2 to n-1 do if n mod i = 0 then ok:=false; if i>=2 then NT:=ok else NT:=false; End; 1.0In cc s nguyn t <= nBegin Write('Nhap so nguyen n: '); Readln(n); Write('Cac so nguyen to khong lon hon n la: '); For i:= 2 to n do if NT(i) then write(i,';'); 0.5S nguyn t nh nht ln hn n. Writeln; i:=n; While not NT(i) do i:=i+1; Write('So nguyen to nho nhat khong be hon n la:',i); 1.0In cc cp s nguyn t l hai s l lin tip. Writeln; Write('Cap cac so nguyen to la hai so le lien tiep:'); For i:=3 to n -2 do if NT(i) and NT(i+2) then writeln(i,';',i+2); Readln; End. 1.0 UBND HUYN QU SN PHNG GD&TK THI HC SINH GII LP 9 CP HUYN NM HC 2010 - 2011 Mn: Tin hc Thi gian lm bi: 150 pht (Khng k thi gian giao ) HNG DN CHM - VNG II Bi 1: ( 6.0 im) Khai boProgram Bai1abc; Var a,aa,b: longint; cs,tong:integer;0.5Hm tm s i xngFunction DX(a:integer):integer; var so: integer; Begin so:=0; Repeat so:=so*10 + a mod 10; a:= a div 10; Until a = 0; DX:=so; End;2.0Tm s ch s ca mt sBegin Write('Nhap so a: '); Readln(aa); a:=aa; cs:=0; Repeat cs:=cs + 1; a:=a div 10; until a=0; Writeln('So a co ',cs,' chu so'); 1.5Tm tng cc ch s ca mt s a:=aa; Tong:=0; Repeat tong:=tong + a mod 10; a:=a div 10; until a=0; Writeln('So a co tong cac chu so la:',tong); 1.5In s i xng v tng i xng a:=aa; Writeln('So doi xung cua so a la: ',DX(a)); Write('Nhap so b: '); Readln(b); Write('Tong doi xung cua hai so a, b la: ',DX(DX(a)+DX(b))); Readln; End. 0.5 Bi 2: ( 4.0 im) Khai boProgram Vong2_Bai2; Var M: array[1..100] of integer; n,i,j:integer; bac: integer;0.5Lp v in tam gic pascalBegin Write('Nhap bac n: ');Readln(n); For i:=1 to n+1 do M[i]:=0; M[1]:=1; Writeln('Tam giac Pascal bac ',n,' :'); For i:=0 to n do Begin For j:=i+1 downto 2 do M[j]:=M[j] + M[j-1]; For j:=1 to i+1 do write(M[j]:4); Writeln; End;2.0In khai trin liy tha bc n ca nh thc. 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