Задача: реализовать метод Тихоновна по решению некорректных задач. пример тут: http://letitbit.net/download/fedbf2800813/-------.doc.html сама прога на паскале: Текст программы для реализации метода Тихонова на языке PASCAL Uses CRT; type real=extended; const matrixA: array[1..3,1..3] of real = ((-19/20,1/5, 3/5), (-1 ,0.1, 0.5), (-0.01 ,0 ,1/200)); One: array [1..3,1..3] of real = ((1,0,0), (0,1,0), (0,0,1)); U:array[1..3] of real = (1,1,1.1); var i,j,k,q:byte; A,At,A1,A2,Ar,One1:array[1..3,1..3] of real; delta,Det,S,alpha:real; B,Z,U1:array[1..3] of real; f:text; Procedure TransA; begin for i:=1 to 3 do for j:=1 to 3 do At[i,j]:=A[j,i] end; Function Koef(par1,par2:byte):real; var Sum:byte; Tmp:real; begin Sum:=par1+par2; Tmp:=1; for k:=1 to sum do Tmp:=Tmp*(-1); Koef:=Tmp; end; Function AlAdd(par1,par2:byte):real; type element=record value:real; flag:boolean; end; var BB:array[1..2,1..2] of real; AA:array[1..3,1..3] of element; k,v,w:byte; N:array[1..4] of real; P1:real; begin for v:=1 to 3 do for w:=1 to 3 do begin AA[v,w].value:=A2[v,w]; AA[v,w].flag:=true end; for v:=1 to 3 do AA[par1,v].flag:=false; for v:=1 to 3 do AA[v,par2].flag:=false; { for v:=1 to 3 do begin for w:=1 to 3 do write(AA[i,j].value:2:3,' '); writeln end; } k:=1; for v:=1 to 3 do for w:=1 to 3 do begin if AA[v,w].flag then begin N[k]:=AA[v,w].value; { writeln(N[k]);} k:=k+1 end; end; BB[1,1]:=N[1]; BB[1,2]:=N[2]; BB[2,1]:=N[3]; BB[2,2]:=N[4]; { writeln('alg dop',par1,par2,' ',BB[1,1]*BB[2,2]-BB[1,2]*BB[2,1]);} AlAdd:=BB[1,1]*BB[2,2]-BB[1,2]*BB[2,1]; end; Function DetCount:real; var S1:real; z:byte; begin S1:=0; for z:=1 to 3 do S1:=S1+A2[1,z]*Koef(1,z)*AlAdd(1,z); DetCount:=S1; end; Procedure RevMatr; begin for i:=1 to 3 do for j:=1 to 3 do Ar[j,i]:=Koef(i,j)*AlAdd(i,j)/DetCount; { for i:=1 to 3 do begin for j:=1 to 3 do write(Ar[i,j],' '); writeln; end;} end; Function AllRight:boolean; begin writeln(f,'*Ґўп§Є Ї® 1-¬г н«-вг',(abs(U[1]-U1[1]))); writeln(f,'*Ґўп§Є Ї® 2-¬г н«-вг',(abs(U[2]-U1[2]))); writeln(f,'*Ґўп§Є Ї® 3-¬г н«-вг',(abs(U[3]-U1[3]))); writeln(F); if (abs(U[1]-U1[1])<0.001) and (abs(U[2]-U1[2])<0.001) and (abs(U[3]-U1[3])<0.001) then AllRight:=true else AllRight:=false end; Function Pow(par1:real;par2:byte):real; var S2:real; z:byte; begin S2:=1; if par2=0 then begin Pow:=1; exit end else for z:=1 to par2 do S2:=S2*par1; Pow:=S2; end; BEGIN clrscr; Assign(f,'c:\tikh.txt'); Rewrite(f); for i:=1 to 3 do for j:=1 to 3 do A[i,j]:=matrixA[i,j]; TransA; Det:=0.000125; {----------------------------} for i:=1 to 3 do begin S:=0; for j:=1 to 3 do begin S:=S+At[i,j]*U[j]; B:=S end; end; {----------------------------} for i:=1 to 3 do for j:=1 to 3 do begin S:=0; for k:=1 to 3 do begin S:=S+At[i,k]*A[k,j]; A1[i,j]:=S end end; {-----------------------------} q:=1; repeat alpha:=q/pow(4,q); for i:=1 to 3 do for j:=1 to 3 do One1[i,j]:=One[i,j]*alpha; for i:=1 to 3 do for j:=1 to 3 do A2[i,j]:=One1[i,j]+A1[i,j]; RevMatr; {------------------------------} for i:=1 to 3 do begin S:=0; for j:=1 to 3 do begin S:=S+Ar[i,j]*B[j]; Z:=S end; end; for i:=1 to 3 do begin S:=0; for j:=1 to 3 do begin S:=S+A[i,j]*Z[j]; U1:=S end end; q:=q+1; until AllRight; {------------------------------} clrscr; writeln('ЏаЁЎ«Ё¦Ґ*ЁҐ Є *®а¬ «м*®¬г аҐиҐ*Ёо'); for i:=1 to 3 do writeln('Z(',i,')=',z); writeln; writeln('‡* зҐ*ЁҐ Їа ў®© з бвЁ ЇаЁ Ї®¤бв *®ўЄҐ ЇаЁЎ«. аҐиҐ*Ёп'); for i:=1 to 3 do writeln('U1(',i,')=',U1); writeln; writeln('‡* зҐ*ЁҐ Ї а ¬Ґва ॣг«паЁ§ жЁЁ:'); writeln(alpha); Close(f); readln; END. Помогите за вознаграждение плиз.
а что не понятно? О_о Code: matrixA: array[1..3,1..3] of real = ( (-19/20,1/5, 3/5), (-1 ,0.1, 0.5), (-0.01 ,0 ,1/200) ); One: array [1..3,1..3] of real = ( (1,0,0), (0,1,0), (0,0,1) ); U:array[1..3] of real = (1,1,1.1); на: Code: double matrixA[3][3] = {{-19/20,1/5, 3/5}, {-1 ,0.1, 0.5}, {-0.01 ,0 ,1/200} }; double One[3][3] = {{1,0,0}, {0,1,0}, {0,0,1} }; double U[3] ={1,1,1.1}; Code: var i,j,k,q:byte; A,At,A1,A2,Ar,One1:array[1..3,1..3] of real; delta,Det,S,alpha:real; B,Z,U1:array[1..3] of real; f:text; Ha: Code: BYTE i,j,k,q; DOUBLE A[3][3],At[3][3],A1[3][3],A2[3][3],Ar[3][3],One1[3][3] DOUBLE delta,Det,S,alpha DOUBLE B[3],Z[3],U1[3] FILE *f
вот структура программи на с++ и пример: Code: #include <iostream> #include <cstring> using namespace std; int func(double param1) { int i = 55; return i; } void main(){ cout<< func(5); } Примери: Code: Procedure TransA; begin for i:=1 to 3 do for j:=1 to 3 do At[i,j]:=A[j,i] end; Ha: Code: void TransA(){ for(i=1;i<3;i++){ for(j:=1;j<3;j++){ At[i][j] = A[j][i]; } } } Code: Function Koef(par1,par2:byte):real; var Sum:byte; Tmp:real; begin Sum:=par1+par2; Tmp:=1; for k:=1 to sum do Tmp:=Tmp*(-1); Koef:=Tmp; end; На: Code: double Koef(BYTE par1,par2){ byte Sum; double Tmp; Sum = par1 + par2; Tmp = 1; for(k=1;k<Sum;k++){ Tmp = Tmp * (-1); } return Tmp; } Дальше заменяй сам) Ничего сложного нету
a где вознаграждение? main: Code: clrscr; for i:=1 to 3 do writeln('Z(',i,')=',z[i]); writeln; for i:=1 to 3 do writeln('U1(',i,')=',U1[i]); writeln; writeln(alpha); Close(f); readln; END. Ha: Code: void main(void){ system("CLS"); for(int i=1;i<3;i++){ stdout>>"Z[",i,"] = ",Z[i]; } for(int i=1;i<3;i++){ stdout>>"U1[",i,"] = ",U1[i]; } stdout>> alpha; fclose(f); } Робота с файлами: Code: FILE *f; Открытие: <поток> = fopen("путь к файлу", "тип") r - открыть для чтения; w - для записи; w+ - для чтения и записи; Ввод и вывод данных в файл getc(); putch(); соответсвенно посимвольно.... fgets(); fputs(); соответсвенно построчно getw(); putw();
