Раскладываем по степеням двойки: 67 (10cc) = 64+2+1 = 2^6+2^1+2^0 = 1 000 011 (2cc) ИЛИ 67 |2 66 | | 33 | 2 1 32 | | 16 |2 1| 16 | | 8| 2 0| 8| ---| 4|2 0| 4| --|2|2 0|2| --| 1 < первая цифра в числе и далее - остатки 0 49 (10cc)=32+16+1=2^5+2^4+2^0=110 001 (2cc) 54= 32+16+4+2= 2^5+2^4+2^2+2^1= 110 110 (2cc) 81=64+16+1=2^6+2^4+2^0=1 010 001 (2cc)
Оксана759
03.08.2022
Const e=0.001; var x,xn,h,an,s,f,y:real; n,k,i:integer; begin n:=10; h:=(1-0.1)/(n-1); x:=0.1-h; for i:=1 to n do begin x:=x+h; an:=1; xn:=1; f:=1; k:=0; s:=1; while an>e do begin xn:=xn*x*x; k:=k+1; f:=f*k; an:=(2*k+1)*xn/f; s:=s+an; end; y:=(1+2*x*x)*exp(x*x); writeln('x = ',x:4:1,' s = ',s:6:4,' y = ',y:6:4); end; end.
Результат: x = 0.1 s = 1.0303 y = 1.0303 x = 0.2 s = 1.1241 y = 1.1241 x = 0.3 s = 1.2911 y = 1.2911 x = 0.4 s = 1.5490 y = 1.5490 x = 0.5 s = 1.9260 y = 1.9260 x = 0.6 s = 2.4653 y = 2.4653 x = 0.7 s = 3.2320 y = 3.2320 x = 0.8 s = 4.3240 y = 4.3240 x = 0.9 s = 5.8894 y = 5.8895 x = 1.0 s = 8.1548 y = 8.1548