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( n" S0 C T$ A5 \+ v% B7 p4 s0 dif((a[3]-a[0])<e) ) X9 _9 D, q" ?5 Z f9 `4 v* j/ w/ C, v" J# w# r4 B
{ aa=(a[1]+a[2])/2; *ff=function(aa);7 m, ]* n: c' ^, f" `; s
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break ;$ i+ }8 w0 B) Y2 b1 J: [ n
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9 a0 F( O+ e+ Kreturn ;6 d5 d6 I9 R% z' T9 _1 R
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void main()1 m4 b$ d9 ^1 ~! A X+ O7 F
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float xx, ff; ' {' ^8 @8 ~* Q8 E" U: n; m+ I5 u0 j0 y/ X
xx=gold(&ff); ( B5 p) W7 Y0 U5 E- g 5 ?" n" A" E# r: P/ k' S! qprintf("\nThe Optimal Design Result Is: \n"); * f7 O: D0 J; r: m8 N 1 |3 M* D7 t& Q% Bprintf("\n\tx*=%f\n\tf*=%f",xx,ff);& v, ]! K- C$ E3 P
5 b& s7 e& P7 A& S$ \. w1 h1 r& d$ Lgetch(); 0 ]- f1 C0 @: Z ]9 \ ~" o ~0 A* Z G
} 4 k. z/ K* F- }2 u# z3 p8 Q . @$ E% ~4 n4 u) H# L" |$ y+ T! l9 m7 z. t% y #include <stdio.h>- [# U2 y4 F9 S! R) o! k #include <math.h> 6 C+ }7 D7 `) w) g, C; G#define e 0.001* B3 a/ W! T: A1 [0 e7 O, N5 W #define tt 0.001 3 I6 ^# D L# A- xfloat f(double x) ! Z* F2 r5 h7 y9 a2 |. [6 C6 O8 X{' F) B( q! b( C float y=9*pow(x,4)-7*pow(x,3)-4*pow(x,2) +2; , G; u6 W- Z$ |2 ~3 Z$ ?return(y); : g" y, y6 a. e9 ]( E$ Y3 F} & ? e) C2 O0 U4 |finding(float *p1,float*p2) - |0 ^3 G- e5 v; V{ 1 B, w' V4 O# }4 l% \4 x8 D! sfloat x1=0,x2,x3,t,f1,f2,f3,h=tt; 7 B0 D0 t; G* o9 r' _int n=0; / V# o5 U4 T- b5 ]x2=x1+h;f1=f(x1);f2=f(x2); : V |( [9 K$ [3 a X% ]if(f2>f1) {h=-h;t=x2;x2=x1;x1=t;} 5 ]) w) K4 v4 {9 k1 M) cdo 7 ^( b/ x' h' ?! b" U5 x5 e5 v' h7 M' O2 Z1 j: |+ {
{ x3=x2+h;h=2*h;f3=f(x3);n=n+1;}. M0 }6 A& H% `) g while(f3<f2);4 W6 y4 U0 B; A% l if(x1>x3) {t=x1;x1=x3;x3=t;} * o3 b) g/ o3 R# b6 |5 d*p1=x1;*p2=x3; $ o' c: u+ c* |& Areturn(n);' M0 O. C# L h6 y. [4 I }6 i4 P1 y9 h& e' r gold(float *p) 7 I' V1 C3 {& W) N# q N. x) k{' p2 F; p5 x; g$ O! r) v float a,b,x1,x2,f1,f2; 6 i. V+ f. R% ^7 Eint n=0;1 c9 r( M8 _7 n1 ^* Y* u finding(&a,&b);4 |! ]* N& e" e do 1 F" A. |( ~% ?1 |$ x J " s0 k' i# Z) O$ a' W{x1=a+0.382*(b-a);$ ]" s( p8 w3 i- o5 R8 ]$ K0 { ; ]) e4 T& S+ B' R6 z: _
x2=a+0.618*(b-a);f1=f(x1);f2=f(x2);n=n+1;5 M: m: S0 m* K9 b8 K# E( E( N - R7 `- j: v4 }
if(f1>f2) a=x1;7 x, I5 E' ?% [5 b& F & E k0 x# D ^# [else b=x2;} 5 Q+ s# K$ N( |4 R- Q& P6 C4 }: o1 A: l$ Gwhile((b-a)>e); & f5 d& C, o& p& v6 U*p=(x1+x2)/2;return(n); ( a; g* t6 u+ f: {- o. x}6 H& S. v( w# S* h* R) J$ @" K
' i3 A; [& f+ N% X- w; f) h * @+ l, `# Q# ^0 K! |6 c& U0 [9 b' I3 l0 U: n- M( n main(): s/ b4 U k9 m. x1 a J8 E- C {7 o+ L9 [& A" U' a' J, x4 E' B float a,b,x,min;int n1,n2; 4 i' Y% c( K6 [ ^7 \) h; V7 In1=finding(&a,&b); , |4 C8 X7 A2 u, Z- `2 G" o8 N; Un2=gold(&x); 0 o5 j( ]/ R/ g$ u; [min=f(x);+ K2 A& V" C5 V( G* z8 ~( a) L printf("\n The area is %f to %f.",a,b); printf("\n The number 1 is %d.",n1); - G( C, Y/ W( Kprintf("\n The min is %f and the result is %f.",x,min);: r' B" w+ t7 X' T printf("\n The number 2 is %d.",n2);}: ]5 o4 S% x2 @
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The area is 0.000000 to 1.290000. 8 m3 Y+ n6 m1 L - u4 Z5 b' l, R" g% M. ?The number 1 is 8. 9 V" j# W7 J* K. p( N2 b5 c ?3 T5 ]" J+ q
The min is 0.845896 and the result is -0.491089.5 G9 _4 n& Y- O+ g9 d' l; ^% d3 x
$ E( i+ R+ ?9 i+ x' X( _; AThe number 2 is 15 9 v, p' B! U: @0 u2 U x$ y