bài tập bài toán chiếc túi xách

12 503 2
bài tập bài toán chiếc túi xách

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

Thông tin tài liệu

TAP.CCT a. Ph5t bidu bdi to5n: Bii tQp 1: Bii toSn chi€'c triix5ch G\/ MOt c6i kho chfa n loai d6 v6t c6 kich thr/6c vd gi5 tri khiSc nhau. Cu thd: Loai db vgt i (i=1 n) c6: - Kich cd m[i] e N. - Gi5 tri c[i] e R - Sd lrtOng: khdng han ch6 M6t t6n tr6m mang theo chi6c t[i c6 kich cd ld p e Nx. V6y h6n ph6i chon lda mdt danh siich ciic db vdt s6 mang di nhd th6 nio dd cho tdng gi5 tri t6y c5p dddc ti t6n nhdt. Tfc: Tim x[1], x[2],,,,, x[n] (vdi x[i] e N : sd lLtdng loai db v6t thf icen ldy) sao cho: ,; Ix[i].mlil < p ve lx[i].c[i] dat gi5 tri ctrc dai x Input: + n ld sd d6 v6t, kich thd6c cdc db khSc nhau. Loai d6 v6t i 1 2 n Kich cd mlil mlll ml2l mlnl Gi5 tri clil cl 1l cl2l clnl Vdi m[i] e N' , c[i] e R + p lA cd trii x Output: Tim (X[1], X[2] . . . X[n]) (X[i ] :sdtoaidbv6t i, i=1 n) fZ,lfulil= p 5ao cno: tI'*t,i.t,t-+ max n Loai db v6t i 1 2 3 4 5 Kich cd mli 5 6 3 7 B Gra tn cl I 2 5 2 6 1 r' Vidt1: n = 5 ; p = 10, dtJdccho x K€t guir: + Sd db virt 5, kich thd6c 1"0 + Tdng gi5 tri nh6p dr/dc vdo t(i xSch: B + Sd lrrOng ldy theo th[r tlr nhd sau: 0 0 1 1 0 b. Ph6n tich bii to6n: Brl6c 1: Ph6n tich bii toSn: Goi P(r, s) li bdi todn chi6c tfi x6ch, vdi: r e Nx: kich cd chi6c t[i s e N*: s6 c5c loai d6 v6t khdc nhau (bdi todn ban d6u td p(p, n)) C5c gi5 tri cEn tim: t[r,s]: gi5 tricdc dai Ixlil,c[i] cia bditoSn p(r, s) [r,s]:_s6 h/dng loaidb v6t s t6i uu c6n l6'y (tr?c: x[s]) ctia bditodn p(r, s) Brfdc 2: ciii phSp ct€ quy: Khis=1: u[r, 1] = r div m[1] Ilr, 1l = u[r, 1]*c[1] Khis>1: l[r, s] = max { k*c[s] + lfr - kxmfs], s-11 ] 0<k<rdivm[s] = koxclsl + lfr - h*mls], s-11 ufr, s] = ks Brldc 3: LQp bing Procedure LapBang; Begin For s:=1 to n do For r:=0 to p do If s=1 then Tinh u[r, 1] vir l[r, 1] else Tfnh u[r, s] vd l[r, s]; End; Vi du s R 1 2 3 4 5 o o\o o\0 o\o 0\0 o\0 1 0\o 0\o o\o o\0 o\o 2 0\0 0\0 0\0 o\0 0\0 3 0\o 0\o 2\1 2\o 2\O 4 o\o 0\o 2U 2\o 2\0 5 2\1 2\O 2\o 2\o 2\o 6 2U s\1 s\o s\o s\o 7 2\1 su s\o 6\1 6\o 8 2U s\1 s\0 6U 6\0 9 2\1 s\1 7\1 7\0 7\0 10 4\2 su 7\1 8U 8\0 x x€'t qui: + Sd d6 vat s, kicfr thrrdc to + Tdng gi6 tr! nh6p dUdc vdo tfixiich: B + Sd luOng ldy theo thf/ tLr nhut sau: 0 0 1 1 0 D0 phrirc tap tinh to6n: O(np2) BrJ6c 4: Tdng hqrp k€t qui Procedure TongHop; Begin r: =pi For s:=n downto 1 do Begin x[s]:= u[r,s]; r:= r - x[s],mfsl; End; End; E6 phr?c t?e tinh to6n: O(n) BrJ6c 5: Cei dit chddng trinh Program TuiXach; Uses crt; Var n,p:word; x,m:arrayf1., 10] of word; c:arrayf1 10] of real; l:array[0 10,1 10] of real; u:array[0 10,1 10] of word; { } Procedure NhapDL; Var i:byte;f:text; Begin assign(f,'tuixach.inp'); reset(f); read(f,n); readln(f,p); End; { } Procedure Lapbang; Var r,s,k,k1 :word; max,tam : real; Begin for s:=1 to n do for r:=0 to p do if s=1 then Begin u[r,1]:= r div m[1]; l[r,1]:=uIr,1]*cl1]; End else Begin k1: =0; max:= 0; for k:=0 to r div m[s] do Begin for i:=1 to n do read(f,mlil); for i:=1 to n do read(f,c[i]); writeln('So vat ',n,', kich thuoc lieu trong file tuixach.inp'); close(f); tui 'rp,', du Tam : = kxclsl +lIr-kxm[s],s-11; if tam>max then Begin Max:=tam; k1:=k; end; end; lIr,s]:=max; u[r,s]:=k1; End; End; { } Procedure tonghop; Var r,s:word;f:text; Begin assig n(f,'tuixach.out'); rewrite(f); r:=pj for s:= n downto 1 do Begin xlsl: =uIr,s]; r:=r-x[s]xm[s]; End; End; { } Procedure Inkq; Var s:word; f:text; Begin assig n(f,'tu ixach.out' ); rewrite(f); writein('Tong gia tri nhan duoc rrao tui xach la: ',1[p,n]:4:1); write('So luong lay theo thu tu nhu sau: '); for s:=1 to n do Begin write(f,xIs]:3); write(x[s]:3); End; writeln; close(f); End; { } Procedure Inbang; Var srr:word; Begin writeln; for s:=1 to n do write(s:S,' '); writeln; for r:=0 to p do Begin write(r); for s:=1 to n do Begin write(lIr,s] : 3 : 0,'\', u Ir,s],"); End; writeln; end; End; t ) BEGIN Clrscr; NhapDL; Lapbang; Tonghop; inbang; Inkq; Readln; END. { } BiitQp 2: ph6p nhAn td hdp nhi€u ma trQn (O", ) \J a. PhSt bidu bii to6n CEn tinh M = M1xM2x ,xMn Trong d6: M, Id ma tr6n c5'p m[i-1]xm[i] (i=1 n) . H5y xdc dinh thf tUthr/c hi6n cdc ph6p nh6n sao cho s6 ph6p tinh Ie t6ithidu. x Input: + C6'p cr]al ma tr6n Vdi m[i] e N x Output: X5c dinh thf td tinh: sd ph6p tinh > min xVidg:TinhM= M1 * MZ " M3 | 10x201[20xs0] [s0xs] (MrxM2)xM, c6 sd ph6p todn td: 10x20x50 + 10x50x5 = 12500 M1x(M2xM3 ) c6 sd ph6p todn ld: 10x20x5+20x50x5=6000 x K€'t qui: M1 x M2 x M, c6 sd ph6p todn lir: 6000 b. PhAn tich bii to6n: Bddc 1: Ph6n tich bii toin: Goi P(r, s) ld bdi todn nhdn ma tr6n: MrxMr*rx xM' vdi r < s cua n ma tr 0 1 n mlil ml0l mlll mlnl (bai todn ban d6u td p(1, n)) Gi5 tric6n tim: !lt,r]t vitri ph6p to5n thrrc hi6n cudi cing cia beitodn p(r, s) (MrxMr*1x xMp-1) x (MpxMp* 1" rM5) k = k[r,s] e [r+1, s] [r, s]: sd ph6p tinh nh6n t6i uu cira baitodn p(r, s) BtI6c 2; ciii ph5p dG quy: Trr/o'ng hdp 1: Khis = r: l[r, r] = g Trr/dng hgp.2t KhL5 = p1g. lt r,Q t\< hf'- r fx m Jf.1 J I q 1 kfqv+11 =-1+1 Trddng hdp 3r Khi s > r<S: l[r, s] = min { l[r, v-1] + m[r-1]*m[v-1]*mlsl + lfv, sl ] r+1 <v<s (v lA c5c vi tri ph6p todn thdc hi6n cudi cirng khdc nhau) = l[r, vL1] + mlr-11 * mlvll] {, m[s] + l[v', s] (v' ld vi trit6i rru trong sd cdc vi tri cira v) k[r, s] = Y' Nh6n X_6t: C6 thd gh6p trddng hdp 2 vAo trddng hdp 3l Brfdc 3: Lflp bing Procedure LapBang; Begin For s:=1 to n do For r:=s downto 1 do If r = s then lfr, r] := 0 End; else Tinh l[r' s] vi k[r' s]; Vi^!U: Tr/ vi du tr6n ta l6p dddc bing sau 56 ph6p to6n tdi thidu td: 6000 Thf t{thrJc hi6n ph6p nh6n: 3 2 r s 1 2 3 1 0lt 2 10000/2 012 3 600012 s000/3 013 D6 phr?c t?p tinh to6n: O(n3) Brf6c 4: Tdng hep k6t quJ Procedure TongHop; Begin writetn('Sd ph6p tinh le t6i thidu:,, l[1, n]); i:=r'ti 1-imW(1, n); {NhEm lEn [fdt xiic dinh cdc gi5 tri x[i]: vi tri ph6p nhSn cEn thuc hi6n trong tEn nh6n thf i(i=1 n-1)) writeln('Thr? t{ thttc hi6n cdc ph6p nh6n :,); For i:=1 to n-1 do writeln(xfil); End; Trong d6 th[r tuc TimW dr/dc x6y dqfng d9 quy nhf sau: Procedure TimW(r, s); Begin If (r < s) then Begin i:=i-1; vt:= k[r, s]; xlil:= vt; TimW(r, vt-1); TimW(vt, s); End; End; Bddc5: CAi d{t chtldng trinh Program Matran; Uses Crt; Var n:word; m :array[0 20] of word; x:arrayf1 20] of word; k,l :arrayf1 20,1 20] of integer; { } Procedure NhapDl; Var i:byte;f:text; Begin assign(f,'matran, inp'); reset(f); readln(f,n); for i:=0 to n do Begin read(f,mIi]); write(mIi]:3); End; writeln; close(f); End; { } Procedure Lapbang; Var r,srvrvmax:word; Begin for s:=1 to n do for r:=s downto l" do if s=r then Begin lIr,r]:=0; klr,rl:=r; End else Begin VmaX:=f + 1; l[r,s]:=l[r,r]+mIr- llxmIr]xm[s] +lIr+ 1,s]; for v:=r+2 to s do if l[r,s]>l[r,v-1]+m[r-1]xm[v- llxm[s]+l[v,s] then Begin VMaX:=V; llr,sl : = lIr,v- 1] +mIr-1] xmlv- llxm[s]+l[v,s]; End; k[r,s]:=vmax; End; End; { } Proced ure timvt( r,s :word) ; Var vt,i:word; Begin if r<s then Begin i:=i-1; vt:=k[r,s]; xlil:=vt; timvt(r,vt-1); timvt(vt,s); End; End; t ) Procedure tonghop; Var i:word;f:text; Begin assign(f,'matra n.out'); rewrite(f); writeln('so phep toan toi thieu la: ',1[1,n]); writeln(f,'so phep toan toi thieu la: ',1[1,n]); i:=rtj timvt(1,n); write('thu tu thuc hien cac phep nhan:'); writeln(f,'thu tu thuc hien cac phep nhan:'); for i:=1 to n-1 do Begin write(xIi]:3); write(f,xIi],' '); End; Close(f); End; { } Procedure Inbang; Var s,r:word; f:text; t Begin writeln; l for s:=1 to n do write(s:10,' '); -j writeln; for s:=1 to n do Begin write(sr' '); for r:=1 to s do t Begin write(lIr,s]:5 ,'f' ,klt,s],' '); End; writeln; end; End; { } BEGIN Clrscr; NhapDL; Lapbang; Tonghop; Inbang; Readln; END. Bii tdp 3: Tim x6u trong cdc dai. a. PhSt bidu bii to6nr Bei toSn x6u t1o1s cr"lc dqi: s ld x6u trong cia T n6u s nh6n dddc bSng cSch xo5 di m6t sd t<f ttl nAo d6. Vidu: 'ABC, ld x6u trong cria.GAHEBOOC, Bei todn: cho 2 x6u T1, T2. Tim m6t x6u S ld x6u trong chung cria T1 vd T2 co d6 ddi ct;c dAi, '-"r -"errJ vev 'r ve Vi du: TI='ABCDAE'vd T2='XYACADK' c6 x6u 'AcD' le x6u trong chung vdi d6 ddi cdc dai. * Input: X6u T1 vd T2 * output: X6u s li x6u trong chung cira T1 vir T2 c6 d6 dei cdc dai b. PhAn tich bii tof n: Budc 1: phin tich bii toin: Goi n lA d6 dei cria x6u T1 m ld d6 ddi c[ra x6u T2 Ta cEn xdc dinh x6u con chung S ddi nh6't c0a 2 x6u T1 vd T2 Gi6 sir L(i,j) c6 d6 dei x5u con chung ldn nh6't crla z x6u T1 c6 d6 dii i vd x6u T2 c6 d6 ddij (i<=n, j<=m). Btt6c2= ciii ph6p d6 quy: + Ndu i=0 thi L(0,1;:9, Ndu j=g thi L(i,0)=0 + Ndu i>0 vd j>O va T1;=72-thi L(ij)'= i + L(i_1j_1) + N6u i>0 vd j>0 vd T1i<>T2j tni L(i;l = rvaxiLli,j_rj,l1i_r,i;; Brf6c 3: L6p bing; Procedure LapBang; Begin For i:=1 to n do L[1,0]; =0; For j:=1 to m do L[O,j]: =0; For i:=1 to n do For j:=1 to m do if T1[i] = T2[] then L[i,j]:= 1 + Lli_1,j_11 else L[i,j] := maxab(L[i,j_1], L[i_1,j]); End; Vi du: cho hai x6u T1='ABCDAE'vd T2-'XyACADK,. Tim x6u con chung c6 d6 dii rdn nh6.t. BA I L. * c.D 4 , G , \T; T1\ J\ A TZ I A a C (n A a D C K l 1 A 0 0 t€ I I 1 t B 0 0 lF I I I I (- 0 0 I oz w'l 2 2 D 0 0 I 1 L 2 3@ T*3 A 0 0 1 2 3 J ?= ,& Kdt qu6 x6u chung ddi nh6't ld: ACD D0 phr?c tap tinh to5n: O(n.m) Brl6c 4: Tdng h{p k6t qui: Procedure Tonghop; Begin i:=hi j:=m; S:="; While (L[i,j]>0) do Begin if T1[i]=T2Dlthen Begin S:=T2Ul + S; :._; {, l,=1;' End else if L[i, j]= Lli-1, jl then i:=i-1 else j: =j- 1; End; Writeln('Xau chung dai nhat:', S); End; D6 phfc tap tinh to6n: O(n). Bd6c 5: Cii det chddng trinh Program XauChung; uses crt; Var L:Array[0 100,0 100] of Byte; s1,s2,kq: String; m,n: Byte; { } Function maxab(a,b: byte) : byte; Begin if a>b then maxab: =a else maxab: =b; End; { } Procedure NhapDL; Var f : Text; Begin Assign(f,'xauchung, inp'); Reset(f); Readln(f, s1); Readln(f, s2); m:=Length(s1); n: =Length(s2); Close(f); End; { } Procedure LapBang; Var i,j: Byte; Begin For i:=1 to n do lIi,0]:=0; For j:=1 to m do ll0,jl:=0; For i:=1 to m do For j:=1 to n do if sllil = s2ljl then L[i,j]:= 1 + L[i-1,j-1] else L[i,j]:= maxab(L[i,j-1] 1,il); End; { } Procedure Tonghop; Var i,j:byte; f:text; Begin Assi gn(f,'xauchung. out' ); Rewrite(f); i: =mi j: =n; kq:="' While (L[i,j]>0) do Begin if s1[i]=s2Ulthen Begin kq:=s2[] + kq; i:=i-1; j:=j-L; End else if L[i, j]= L[i, j-1] then , L[i- j:=j-1 else i:=i-1; End; Writeln(f,'s1 - ',s1), Writeln(f,'s2 = ',s2); Writeln(f,'Xau chung dai nhat;', kq); Close(f); End; { } Procedure Inkq; Begin Writeln('s1 = ',sl)i Writeln('s2 - ',s2), Writeln('Xau chung dai nhat:', kq); End; t ) i Procedure Inbang; var i,j:byte; Begin \ writeln(' LAp BANG \; write("); for j:=1 to m do write(s2[]:4); writeln; for i:=1 to n do begin wrlte(s1[i]); for j:=1 to m do write(lli,jl:4); writeln; end; End; { } BEGIN Clrscr; NhapDL; Lapbang; Tonghop; Inbang; Inkq; Readln; END. \ a. phSt bidu bii to6n: Bii tQp 5; Bii to5n ngr/di du lich M6t ngfdi di tt/ thdnh phd 0 ddn thdnh ph6 n vd c6 thd di qua n-l thanh phd kh5c 1, 2,.,., n- 1, theo 16trinh: 0-+ i1 - i2 _+ i;,-+ n, trong d6: O < t i t. ik. n, Gi5 v6 c0a xe ditr/thdnh phd id6n thdnh ph6 j ta cgi,i1. Tim m6t 16 trinh ttJ thdnh phd 0 d6n thdnh pr'ro n-rul"."r.)o tdng chi phf v€ gi5 v6 dat cdc tidu,* Input: +:- ld thtinh ph6xudt phdt lrrr vrrr vs vro + s li dfch + Ma tr6n cli,jl (i,j =0 n) tuu trii chi phf x output: L6 trinh ngSn nh6't trr thdnh prr6 o o8n tnirnh phd n. b. Ph6n tich bii todn: Brl6c 1: Ph6n tich bii to6n: x Goi P(r,s) lA bei todn du lich, vdi: - r e N: dinh xudt phdt - s e N: dinh cEn d6n (bei todn ban dEu: p(0,n)) x C5c gi5 tri c6n tim: - l[l,s] : chi phi nhd nh6't d6 di ti/ r _> s c0a bei todn p(r,s). - ufr,sl : dinh k6 cu6i tr6n dL/dng di tf r-> s cira odi toJn p(r,s). BtI6c2= Giii ph6p d€ qui: -Khir=s: l[r's] = 6 u[r,s] = -1 -Khi r + s: l[r,s] = min(l[r,k] + c[k,s]) = l[r,ko]+c[ko,sJ (r-<k<s) ulr,s] = ko Brfdc 3: LAp bing: Procedure Lapbang; Begin Fors:=0tondo Forr: =0tosdo Ifr=sThen Begin l(r, r): = 0; u(r, r) = -1' End Else Tinh l(r,s)) vir u(r,s); End; Dfdng di tU thenh phd 0 ddn thdnh ph6 4 c6 t6 trinh: 0 >1 >2 >4 Chi phi nh6 nhdt dd di tiJ thdnh phd 0 ddn thdnh ph6 4: 15 D6 phr?c t?p tinh to6n: O(n3) Brl6c 4: Tdng hdp: Procedure Tong hop(d,c :word ); Begin i:=0; x[0]:=c; k:=ci while k<>d do Begin i:=i+1i xlil:=u[d,k]; k:=u[d,k]; End; End; D6 phr?c tap tinh to5n: O(n) Brl6c 5: Cii d{c chr.tdng trinh Program DuLich; Uses cft; Const max=100; Var n:word; x:array[0 max] of integer; c,l,u:array[O max,O max] of integer; { } Procedure NhapDL; Var i,j:byte;f:text; Begin assign(f,'C: \d u lich. inp'); reset(f); readln(f,n); for i:=0 to n do for j:=0 to n do read(f,cIi,j]); close(f); End; { i Procedure Lapbang; Var r,s,k,k0:word; Begin for s:=0 to n do for r:=0 to s do if s=r then Begin lIr,s]:=0; u Ir,s]: =- 1; End else Begin k0: =s-1; lIr,s] : = lIr,s-1] +c[s- 1,s]; for k:=r to s-1 do if l[r,s] > l[r,k]+c[k,s] then Begin k0: = k; lIr,s] : = lIr,k] +c[k,s]; End; u[r,s]:=k0; End; End; { } Procedure tonghop(d,c:word); Var i,k,j,t:word;f:text; Begin assig n(f,'C: \dulich.out'); rewrite(f); i:=0; s r o 1 2 3 4 0 o -1 1 7 0 o -1 2 L2 1 5 1 0 -1 3 L2 0 7 1 4 2 0 -1 4 15 2 8 2 3 2 4 3 o -1 x[0]:=c; k:=c; while k<>d do Begin i:=i+1; x[i]:=u[d,k]; k: =u[d,k]; End; . writeln(f,'Duong di tu Tp ', d,, den Tp ',c,, chi nho nhat la:'); for j:=i downto 0 do . Begin write(f,x[]); if j>0 then write(f,' >,); End; writeln(f); write(f,'chi phi la: ',1[d,c]); close(f); End; { } Procedure Inkq; Var f:text; ch:string; Begin Assign(f,'C: \DuLich.Out'); reset(f); readln(f,ch);write(ch); readln(f,ch);writeln(ch); read I n(f,ch);write(ch); close(f); End; { } Procedure InBang; Var i,j:Byte; Begin phi for i:=0 to n do write(i:g);writeln; for j:=0 to n do Begin write(;':4,' '); for i;=0 to j do write(uIi,j],'/,,1[i,j],' '); writeln; End; End; t ) BEGIN Clrscr; NhapDL; Lapbang; Tonghop(0,4); Inbang; Inkq; Readln;END. a. phSt bidu bai toSn: Bia tip 5: Bii toSn sinh vi6n 6n thi, Bei todn sinh vi6n 6n thi: M6t sinh.vi6n cdn T ngdy dd 6n thi n m6n. Theo kinh nghi6m cia anh ta, ndu 6n m6n j trong i ngdy thi drdc oidr'H J1i,;1. cia stJ cho nia .a. a[i,j] (vdi a[i,j] < =a[i+ 1,j]). Tim b6 xUl (sd ngdy 6n m6n j, v6i j=1,.n) sao cho Ix[1=, vd sinh vi€n dat tdng didm idn nhdt (IalxUl, jl-+max). x Input: + n ld sd m6n + m td s6 ngdy + aIij] didm crla 6n thi m6n j trong i ngiry (ali,jl<=a[i+1,j]) x out put: -,, r"J,]rJiT;';f; i'r,=I n) *o *orx[]=, va sinh vi6n dat tdns didm rdn nhdt (IalxUl, jl-+max), b. Ph6n tich biito5n: Brldc 1: ph6n tich bii to6n: x Goi p(r,s) ld bii toiin SV 6n thi r m6n trong s ngEy (bai to6n ban d6u s6 ld : p(n,m)) " * Cdc gi5 tri cEn tim lir: bait"drl[i;rrj.tunn sd didm ldn nh6't mir sinh vi6n c6 thd dat dudc khi 6n thi r m6n trons s ngdy cria _ ufr,gJ : 56 ngdy tdi uu cEn d6 6n m6n r c0a bii todn p(r,s). Bttdc2: ciii phip dQ qui: -flrif = 1:l[1,r] =a[s,1], u[1,s] = I -Khi i> 1:l[r,s] = max(attit it;r-r,s_t1; ) 'fi<t<-sl a[h,r] + llr-1, s-ksl Brf6c 3: L6p bing: u[r's] = ko Procedure Lapbang; Begin for s:=1 to m do for r:=0 to n do End; if (s =.1) then Tinh lf1,sl vd ufl,sl else Tinh l[r,s] vi u[r,s] vf du: c6 3 m6n thi 6n trong 5 ngdy vir didm cdc m6n tr/dng r?ng cdc ngdy nhf sau: . v6t i 1 2 3 4 5 Kich cd mli 5 6 3 7 B Gra tn cl I 2 5 2 6 1 r' Vidt1: n = 5 ; p = 10, dtJdccho x K€t guir: + Sd db virt 5, kich thd6c 1"0 + Tdng gi5 tri nh6p dr/dc. vdi: r e Nx: kich cd chi6c t[i s e N*: s6 c5c loai d6 v6t khdc nhau (bdi todn ban d6u td p(p, n)) C5c gi5 tri cEn tim: t[r,s]: gi5 tricdc dai Ixlil,c[i] cia bditoSn p(r,. Output: X5c dinh thf td tinh: sd ph6p tinh > min xVidg:TinhM= M1 * MZ " M3 | 10x201[20xs0] [s0xs] (MrxM2)xM, c6 sd ph6p todn td: 10x20x50 + 10x50x5 = 1 250 0 M1x(M2xM3

Ngày đăng: 02/06/2015, 10:39

Từ khóa liên quan

Tài liệu cùng người dùng

  • Đang cập nhật ...

Tài liệu liên quan