Mot so de thi Pascal doc

DE THI PASCAL HP1 Director by:Luu Cong Hoan BT 2.29.Giai PT ax+b>0 Program BT229; Uses crt; Var a,b,x:real; Begin clrscr; writeln('Giai BPT dang ax+b>0'); write('Nhap vao a,b= '); readln(a,b); if a=0 then if b>0 then writeln('Pt co vo so nghiem!') else write('Pt vo nghiem!'); if a<>0 then begin if a>0 then writeln('Pt co nghien la x> ',-b/a:0:4); if a<0 then writeln('Pt co nghien la x< ',-b/a:0:4); end; readln End. BT 2.20.Giai BPT bac hai : ax^2+bx+c >0 Uses crt; Var a,b,c,d,x1,x2:real; Begin clrscr; writeln('Giai bpt ax^2+bx+c>0'); write('Nhap vao a,b,c:'); readln(a,b,c); d:=b*b-4*a*c; if a>0 then begin if d<0 then writeln('BPT co vo so nghien thuc'); if d=0 then writeln('BPT co nghien la moi x khac ',-b/(2*a):0:2); if d>0 then begin x1:=(-b-sqrt(d))/(2*a); x2:=(-b+sqrt(d))/(2*a); writeln(' BPT co nghiem x<',x1:0:2,' x>',x2:0:2); end; end; if a<0 then begin if d<=0 then writeln('BPT vo nghien!'); if d>0 then begin x1:=(-b+sqrt(d))/(2*a); x2:=(-b-sqrt(d))/(2*a); writeln('BPT co nghiem la: ',x1:0:2,' <x< ',x2:0:2); end; end; if a=0 then begin if b=0 then begin if c>0 then writeln('BPT co vo so nghiem x thuoc R') else writeln('BPT vo nghiem'); end; if b>0 then writeln('BPT co nghiem x >',-c/b:0:2); if b<0 then writeln('BPT co nghiem x <',-c/b:0:2); end; readln End. BT 2.21.Giai PT bac 4: ax^4+bx^2+c =0 (a#0) Uses crt; Var a,b,c,d:real; t1,t2:real; Begin clrscr; repeat writeln('Giai pttp ax^4+bx^2+c=0 (a#0)'); write('nhap he so a,b,c='); readln(a,b,c); until a<>0; d:=b*b-4*a*c; if d<0 then writeln('pt vo nghiem'); if d=0 then begin if a*b>0 then writeln('pt vo nghiem'); if a*b=0 then writeln('pt co 1 nghiem x=0'); if a*b<0 then writeln('pt co 2 nghiem x1=',-sqrt(-b/2/a):0:2,'; x2=',sqrt(-b/2/a):0:2); end; if d>0 then begin t1:=(-b-sqrt(d))/2/a; t2:=(-b+sqrt(d))/2/a; if a>0 then begin if t2<0 then writeln('pt vo nghiem'); if t2=0 then writeln('pt co 1 nghiem x=0'); if t2>0 then begin if t1<0 then writeln('pt co 2 nghiem x1=',-sqrt(t2):0:2,'; x2=',sqrt(t2):0:2); if t1=0 then writeln('pt co 3 nghiem x1=0; x2=',-sqrt(t2):0:2,'; x3=',sqrt(t2):0:2); if t1>0 then begin writeln('pt co 4 nghiem x1=',-sqrt(t1):0:2,'; x2=',sqrt(t1):0:2); writeln(' x3=',-sqrt(t2):0:2,'; x4=',sqrt(t2):0:2); end; end; end; if a<0 then begin if t1<0 then writeln('pt vo nghiem'); if t1=0 then writeln('pt co 1 nghiem x=0'); if t1>0 then begin if t2<0 then writeln('pt co 2 nghiem x1=',-sqrt(t1):0:2,'; x2=',sqrt(t1):0:2); if t2=0 then writeln('pt co 3 nghiem x1=0; x2=',-sqrt(t1):0:2,'; x3=',sqrt(t1):0:2); if t2>0 then begin writeln('pt co 4 nghiem x1=',-sqrt(t1):0:2,'; x2=',sqrt(t1):0:2); writeln(' x3=',-sqrt(t2):0:2,'; x4=',sqrt(t2):0:2); end; end; end; end; readln End. BT 2.22.Giai PT bac nhat 2 an : ax by c dx ey f + =   + =  Var a,b,c,d,e,f,DT,Dx,Dy:real; Begin writeln(‘Giai HPT bac nhat 2 an so’); write(‘Nhap cac he so a,b,c=’); readln(a,b,c); write(‘Nhap cac he so d,e,f=’); readln(d,e,f); DT:=a*e-b*d; Dx:=c*e-b*f; Dy:=a*f-c*d; if DT=0 then Begin If Dx=Dy=0 then writeln(‘HPT co vo so nghiem!’); If (Dx<>0) or (Dy<>0) then writeln(‘HPT vo nghiem!’); End; if DT<>0 then writeln(‘HPT co ng ! x=’,Dx/DT:0:2,’ y=’,Dy/DT:0:2); readln End. BT 2.23. Cho 3 so thuc duong a,b,c. a, Ton tai hay khong, mot tam giac nhan chung lam 3 canh. b, Neu ton tai, xet tam giac do la vuong, nhon hay tu. Var a,b,c:real; Begin write('Nhap vao 3 so bat ki a,b,c='); readln(a,b,c); if (a>0) and (b>0) and (c>0) and (a+b>c) and (c+b>a) and (a+c>b) then begin writeln('Day la 3 canh cua mot tam giac'); if (a*a+b*b=c*c) or (a*a+c*c=b*b) or (c*c+b*b=a*a) then writeln('Day la tam giac vuong') else begin if (a*a+b*b>c*c) and (a*a+c*c>b*b) and (c*c+b*b>a*a) then writeln('Day la tam giac nhon') else writeln('Day la tam giac tu'); end; end else writeln('Day ko la 3 canh tam giac'); readln End. BT 2.26. Cho so tu nhien n(n<=1000); a, Hoi n co bao nhieu chu so. b, Tinh tong cac chu so cua n. c, Tim chu so dau tien, cuoi cung cua n. Var n,:word; Scs,Tcs,Csd,Csc:byte; Begin repeat writeln(‘Cho so tu nhien n(n<1000)); readln(n); until (n>0) and(n<1000); Csc:=n mod 10; Tcs:=0; Scs:=0; repeat Scs:=Scs+1; Tcs:=Tcs+(n mod 10); Csc:= n mod 10; until n=0; writeln(n, ‘co’,Scs,’so chu so’); writeln(‘Tong cac chu so la’,Tcs); writeln(‘Chu so dau la’,Csd,’;Chu so cuoi’,Csc); readln End. BT 2.29. Nhap vao 1 day cac so nguyen tu ban phim cho den khi gap so 0, Sau do tinh tong cac so duong, trung binh cong cac so am. Var n,Tsd,Tsa:integer; d:byte; Begin writeln('Nhap vao 1 day cac so nguyen cho den khi gap so 0'); Tsd:=0; Tsa:=0; d:=0; repeat write('Nhap n='); readln(n); if n>0 then Tsd:=Tsd+n; if n<0 then begin Tsa:=Tsa+n; inc(d); end; until n=0; writeln('TongSoDuong la:',Tsd); if d>0 then writeln('TrungBinhCong so am la:',Tsa/d:0:2) else writeln(‘Khong co so am nao’); readln End. BT 2.35. Cho so tu nhien h(1<h<25).Su dung ki tu ‘*’ in ra ma hinh tam giac can dac voi chieu cao h Var i,j,h:byte; Begin writeln('In Tamgiac can boi * chieu cao h(1<h<25)'); repeat write('Nhap chieu cao h(1<h<25):'); readln(h); until (h>1) and (h<25); for i:=1 to h do begin gotoxy(41-i,i); {41 48} for j:=1 to 2*i-1 do write('*'); writeln; end; readln; End. BT 2.37. Btoan :”Tram trau tram co, trau dung an 5, trau nam an 3 lai nhai nghe hoa, ba con 1 bo” Var d,n,ng:byte; SoNg:word; Begin SoNg:=0; writeln('Trau dung Trau nam Nghe hoa'); for ng:=1 to 100 do for n:=1 to 34 do for d:=1 to 20 do if 5*d+3*n+ng/3=100 then begin writeln(d:4,n:4,ng:4); inc(SoNg); end; if SoNg>0 then writeln('So nghiem Btoan la',SoNg) else writeln(‘PT vo nghiem’); readln End. BT 2.38. Tim nghiem ngduong PT:2x+5y=k, voi k nhap tu ban phim(k<=10000). Var k,x,y,SoNg:word; Begin writeln('Giai pt 2x+5y=k,k nhap tu ban phim'); repeat write('Nhap so tu nhien k(k<10000):'); readln(k); until (k>0) and (k<10000); SoNg:=0; for x:=1 to Trunc(k/2) do for y:=1 to Trunc(k/5) do if 2*x+5*y=k then begin write('(',x,',',y,') '); inc(SoNg); end; writeln; if SoNg>0 then writeln('So nghiem nguyen duong PT la:',SoNg) else writeln(‘PT vo nghiem nguyen duong’); readln End. BT 2.39. Mot nguoi gui vao ngan hang so tien A dong.Lai xuat moi thang la 0.8%. De co B dong(B>A), can phai gui it nhat bao nhieu thang ? Const LS=0.008; Var A,B:real; t:word; Begin writeln('BToan gui tien tiet kiem'); write('Co bao nhieu?'); readln(A); write('Muon co bao nhieu?'); readln(B); t:=0; while A<B do begin A:=A*LS; inc(t); end; writeln('Can gui it nhat la ',t,' thang'); writeln('Bam phim Enter de ket thuc!'); readln End. BT 2.40. Mot nguoi gui vao ngan hang so tien A dong ,loai co ki han 3 thang (tron 3 thang tinh lai 1 lan) voi lai suat moi thang la 1.0%. Hoi sau t thang nguoi do nhan duoc bao nhieu tien? Const LS=0.01; Var A: real; i,t: word; Begin writeln('Lai suat gui tiet kiem co ky han'); write('Co bao nhieu?'); readln(A); write('Gui trong bao lau(thang):'); readln(t); for i:=1 to Trunc(t/3) do A:= A+3*0.01*A; writeln('So tien co sau ',t,' thang la:',A:0:0); readln End. {Sau du 3 thang moi duoc tinh lai, luc do lai suat duoc nhap vao von, nghia la sau du 3 thang co A:=A+3*0.01*A. so lan duoc tinh la[t/3]} BT 2.44. So n!! duoc dinh nghia nhu sau : 1.3.5 n (n-le) n!! = 2.4.6 n (n-chan)    Cho n thuoc.Tinh tong S= 1!!-2!!+….+(-1) n+1 n!!. Var i,n:integer; S,S1,S2:real; Begin repeat write('Nhap vao so tu nhien n:');readln(n); until n>0; writeln('Tinh S=1!!-2!!+ +[(-1)^(n+1)]*n!!'); S:=0; S1:=1; S2:=1; for i:= 1 to n do if i mod 2 <> 0 then begin S1:=S1*i; S:=S+S1; end else begin S2:=S2*i; S:=S-S2; end; writeln('Tong can tim la S=',S:0:2); writeln('Bam phim Enter de tro ve!'); readln; End. BT 2.48.Cho 2 so nguyen n,m(n,m< 2147483648).Tim (m,n)?[n,m]? Var m,n,mn:longint; Begin writeln('Tim UXLN,BCNN cua 2 so nguyen'); write('cho m,n:'); readln(m,n); if (m=0) and (n=0) then writeln('Ko ton tai UCLN,BCNN') else begin m:=ABS(m); n:=ABS(n); mn:=ABS(m*n); if m*n =0 then writeln('UCLN=',m+n,' BCNN ko ton tai') else begin while m<>n do if m>n then m:=m-n else n:=n-m; writeln('UCLN=',m,' ;BCNN=',mn div m); end; end; writeln('Ban phim Enter de tro ve !'); readln End. {Cach2 : Function UCLN(a,b:longint): longint; var r: longint; begin a:=Abs(a) ; b:=Abs(b) ; while b<>0 do begin r:=a mod b; a:=b; b:=r; end; UCLN:= a; end; } BT 2.49.Cho so tu nhien n(n<214783648).Ktra xem n co thuoc day Fibonaci hay khong? Neu co thi nam o vi tri nao? Var n,f0,f1,f,p:Longint; ok:Boolean; Begin writeln('Ktra n thuoc day Fibonaci?'); repeat write('Cho so tu nhien n(n<2147483648)'); readln(n); until n>0; if n=1 then writeln('So 1 thuoc day, o vi tri 1 hoac 2') else begin f0:=1; f1:=1; f:=2; ok:=false; p:=2;{so n chua thuoc day,Vitri neu co tu 3} while (f<=n) and (not ok) do begin f0:=f1; f1:=f; f:=f0+f1; ok:=(f=n); inc(p); end; if ok then write('So ',n,' thuoc day,o vi tri ',p) else writeln('So ',n,' khong thuoc day Fibonaci'); end; writeln; writeln('Bam phim Enter de ket thuc!'); readln End. BT 2.50. Day so {a k } duoc xac dinh nhu sau: a 1 =1, a k =a k-1 +(k-1), k=2,3,… cho so tu nhien n(n<32768), in ra man hinh cac gtri a 1 ,a 2 ,…,a n . uses crt; Var a,n,k:integer; Begin clrscr; repeat write('Nhap so tu nhien n(n<32767)'); readln(n); until n>0; writeln('Day so can tim la:'); write(1); a:=1; for k:=2 to n do begin a:=a+(k-1); write(' ',a); end; readln End. BT 2.54. So tu nhien a 1 a 2 a k duoc goi la so Amstrong neu: a 1 a 2 a k =a 1 ^k+a 2 ^k+ a k ^k .Tim cac so Amstrong co it hon 4 chu so? Uses crt; Var i,a,b,c:byte; d:word; Begin clrscr; writeln(‘So a 1 a 2 a k , la so Amtrong neu: a 1 a 2 a k =a 1 ^k+a 2 ^k+ a k ^k); writeln('Cac so Amstrong co it hon 4 chu so:'); d:=9; for i:=1 to 9 do write(i,' '); for a:=1 to 9 do for b:=0 to 9 do if 10*a+b=a*a+b*b then begin write(a,b,' '); inc(d); end; for a:=1 to 9 do for b:=0 to 9 do for c:=0 to 9 do if 100*a+10*b+c=a*a*a+b*b*b+c*c*c then begin write(a,b,c,' '); inc(d); end; writeln; writeln('Co ',d,' so Amstrong co it hon 4 chu so!'); readln End. BT 2.55.So n thuoc N duoc goi la so palindrome neu no doi xung(Vd:121,3223, ) Kiem tra so n(n<2147483648) co la so palindrome hay khong? Var d,l,i:integer; n:longint; s: string; Begin repeat write('cho n=');readln(n); until (n>0); str(n,s); l:= length(s); d:=0; for i:= 1 to l div 2 do if s[i] = s[l-i+1] then inc(d); if d = l div 2 then writeln(n,' la xau doi xung') else writeln(n,' ko la xau doi xung'); readln; End. BT 3.4.Cho 3 so thuc a,b,c tinh gia tri cua bthuc. Uses crt; Var a,b,c:real; Function Min(a,b:real):real; begin if a<b then min:=a else min:=b; end; Function Max(a,b,c:real):real; var m:real; begin m:=a; if m<b then m:=b; if m<c then m:=c; max:=m; end; Begin {Ctrinh chinh} clrscr; writeln('Tinh gia tri cua bieu thuc:'); writeln('(min(a,a+b)+min(a,b+c))/(1+max(a+bc,1,2008))'); write('Cho 3 so thuc a,b,c: '); readln(a,b,c); writeln('Gia tri cua bthuc da cho la:'); writeln((min(a,a+b)+min(a,b+c))/(1+max(a+b*c,1,2008))); readln End. BT 3.5.Cho so thuc s,t tinh gia tri bthuc. Var s,t:real; Function H(a,b:real):real; begin H:=a/(1+b*b)+b/(1+a*a)-(a-b)*(a-b)*(a-b);; end; Function Max(a,b,c:real):real; var m:real; begin m:=a; if m<b then m:=b; if m<c then m:=c; Max:=m; end; Begin {Ctrinh chinh} writeln('Tinh gia tri bieu thu'); writeln('P=H(s,t)+Max((h(s-t,st))^2,(H(s-t,s+t))^4,(h(1,1)'); write('Cho 2 so s,t:'); readln(s,t); writeln('P= ',H(s,t)+Max(H(s-t,s*t)*H(s-t,s*t),H(s-t,s+t)*H(s-t,s+t)*H(s-t,s+t)*H(s-t,s+t),h(1,1)):0:2); readln; End. BT 3.6.Tim UCLN cua 4 so nguyen a,b,c,d nhap tu ban phim? Var a,b,c,d:longint; Function ucln(a,b:longint):longint; var r:longint; begin while b<>0 do begin a:=Abs(a); b:=Abs(b); r:=a mod b; a:=b; b:=r; end; ucln:=a; end; Begin write('nhap vao a,b,c,d:'); readln(a,b,c,d); if (a=0) and (b=0) and (c=0) and (d=0) then writeln('UCLN ko ton tai!') else begin writeln('UCLN(a,b,c,d)=',ucln(ucln(ucln(a,b),c),d)); end; readln; End. BT 3.7.Viet ham de quy tim UCLN 2 so nghuyen duong a,b.Ap dung tim UCLN(a 1 ,a 2 , a n ) uses crt; Var i,n:word; a1,u,b:longint; Function ucln(a,b:longint):longint; begin a:=Abs(a); b:=Abs(b); if b=0 then ucln:=a else ucln:=ucln(b,a mod b); {if a*b=0 then ucln:=a+b else if a>b then ucln:=ucln(a mod b,b) else ucln:=ucln(a,b mod a); } end; Function bcnn(a,b:longint):longint; begin a:=Abs(a); b:=Abs(b); bcnn:=a*b div ucln(a,b); end; Begin {Ctrinh chinh} clrscr; writeln('Tim UCLN,BCNN cua n so nguyen nhap tu ban phim'); write('Cho so phan tu n= '); readln(n); write('So thu 1:'); readln(a1); u:=a1; b:=a1; for i:=2 to n do begin write('So thu ',i,':');readln(a1);{nhap so thu i vao a1} u:=ucln(u,a1);{gan u bang UCLN cua u va a1} b:=bcnn(b,a1);{gan b bang BCNN cua b va a1} end; writeln('UCLN= ',u); writeln('BCNN= ',b); readln; End. [...]... s[i]=s[l-i+1] then d:=d+1; Dx:=(d=l div 2); end; Begin writeln('Tim so Hp_ngto giau m va n'); repeat write('Cho m,n(m . ‘co’,Scs, so chu so ); writeln(‘Tong cac chu so la’,Tcs); writeln(‘Chu so dau la’,Csd,’;Chu so cuoi’,Csc); readln End. BT 2.29. Nhap vao 1 day cac so nguyen tu ban phim cho den khi gap so 0, Sau do. ',d,' so Amstrong co it hon 4 chu so! '); readln End. BT 2.55 .So n thuoc N duoc goi la so palindrome neu no doi xung(Vd:121,3223, ) Kiem tra so n(n<2147483648) co la so palindrome. then begin write(i,' '); inc(dem); end; if dem=0 then writeln('Ko co so Hp-ngto nao'); readln; End. Link đề thi HSG Toán +http://dethi.violet.vn/present/show/entry_id/3493381 +http://violet.vn/thcs-thachlap-thanhhoa/present/show/entry_id/5867134 +http://diendan.hocmai.vn/showthread.php?t=185307 +http://www.docstoc.com/docs/7532523/%C4%90%E1%BB%80 -THI- H%E1%BB%8CC-SINH-GI %E1%BB%8EI-L%E1%BB%9AP-8