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if((a[3]-a[0])<e)& P% b& B$ x3 @0 l
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{ aa=(a[1]+a[2])/2; *ff=function(aa);" E; l4 j5 c8 t+ F
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break ; 5 m0 D3 z @6 | v: N" |; \( l * R' P% j7 p: i. ^- R2 ^ w2 w: }4 A$ t% a
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} & S# c5 V( C# n5 A# t' u # s) t. b5 f0 w7 q# D& kreturn ;6 \$ v7 q0 H. L( {) F/ H& ~9 z
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} . ^' t9 n* y" I y7 K + Y" o% ?! j: y- Z' @+ z* pvoid main(): A& p2 `! U( _; }& @8 K, T
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{ 0 h7 s5 r1 u' _$ Jfloat xx, ff; ( }3 d- k k' K; E- z7 H" i 8 h$ p1 m* e$ a$ z4 ?xx=gold(&ff);/ i% _' d1 @1 P6 I
$ {- {& l% K% G8 D( I: k0 z
printf("\nThe Optimal Design Result Is: \n");! y9 \$ d9 [& k( W
2 b! U: O# Y6 k& P$ ~3 I6 U; ~) Aprintf("\n\tx*=%f\n\tf*=%f",xx,ff); 0 a9 V5 \. b' T! m- h1 {" j' k2 U: f$ {
getch(); 8 T1 I0 B- v9 b; p2 o* b# S # Y! T" `% y/ j1 D% l- j} ; ^7 o, G; q& y9 ^6 P& y, u" Q) f2 o' n, b8 J
6 I( D3 l! Q' v #include <stdio.h>+ Z5 }3 M4 _+ p8 _! ^ @) u0 [. E #include <math.h> ' {- d# Z! z8 W% G; ^#define e 0.0019 _9 h0 z: [7 @) Z8 D #define tt 0.001 ( E3 t% O* w6 M) F- ]$ {float f(double x)" J" a1 F& }9 \- b/ }9 m { ' A L% y& N* y5 }. j. ?+ [% e! _float y=9*pow(x,4)-7*pow(x,3)-4*pow(x,2) +2; 2 r5 v3 G% ^& v% ?6 Mreturn(y); 2 O) V- n" F! K" i. Q}$ p& T+ Z' b" ]: F/ ?$ O finding(float *p1,float*p2)/ k! c9 f) g* N9 X" ` {1 w- ]# _" ]; x5 r- D, I: O float x1=0,x2,x3,t,f1,f2,f3,h=tt; ) K- i5 J6 K. v) ?" Q6 e( E+ Fint n=0; t9 Q* y5 C% w; G x2=x1+h;f1=f(x1);f2=f(x2); 9 T5 {* R( q0 ^; Nif(f2>f1) {h=-h;t=x2;x2=x1;x1=t;}* @6 ]+ D/ U) m/ m; h, \; L do8 A' R+ N) |& D: `$ T 8 p, d5 X5 A7 p W9 n{ x3=x2+h;h=2*h;f3=f(x3);n=n+1;}( d$ \3 o3 A4 u( J while(f3<f2);- ?# A5 [1 p: w9 t" j! @: D# S if(x1>x3) {t=x1;x1=x3;x3=t;} 8 x9 O: e) z c) s/ y: D$ f*p1=x1;*p2=x3;; x3 V6 e) y* W" k. e0 f8 Z; | return(n);2 o0 T) A2 K$ h$ Z% G2 ] } . {4 r: N" a3 M6 [gold(float *p), c+ x6 P: y; w* ~' i4 U { s* ]& R2 K* H' t float a,b,x1,x2,f1,f2; 8 ?! ~# N- R1 j5 Jint n=0;9 r) c) c/ Z7 \$ F. O4 G finding(&a,&b);1 T R9 p* d7 P9 S) q' \6 [- a do ' V& {) Q. N' g# A" |1 _: F$ d % o1 g& i( E$ s. @. P [1 Q/ a{x1=a+0.382*(b-a);$ z( l" w6 d9 Y3 ]* A0 w& L ! {3 d/ i; R& T; Nx2=a+0.618*(b-a);f1=f(x1);f2=f(x2);n=n+1; ; L2 D; ?; ?( z* i. z$ t; H& d& U3 w; V8 I, K1 m
if(f1>f2) a=x1;9 y* q( Q. f# q9 f, e # x% k5 u* Q0 l+ v. p; T( w
else b=x2;}# c! I8 Z& l% \6 _3 e1 X while((b-a)>e); 8 v9 q. Z: Y0 }2 p' V1 k0 M6 a- H9 P*p=(x1+x2)/2;return(n); . D" u3 [1 m9 D2 Q0 ?}0 t Q3 {2 Y, A
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! v+ @' H: U. C1 E 8 P# P$ O+ R1 T( j: Amain() , w$ w' K* \$ U- A: y! ~( }4 u{$ D# i/ u4 q5 C6 _' }! X; W% d! A- K float a,b,x,min;int n1,n2; ( m8 m$ t6 \- H0 x' [! Dn1=finding(&a,&b);5 ^2 ]0 Y; i3 ^; J n2=gold(&x); + j" s) @* r( ?min=f(x); |$ ~/ v4 X7 K% l* C printf("\n The area is %f to %f.",a,b); printf("\n The number 1 is %d.",n1); & l& h6 K# A) t: ?3 U& Nprintf("\n The min is %f and the result is %f.",x,min); . B* C7 n0 \2 i( H6 T8 m- u; A8 w% P. Kprintf("\n The number 2 is %d.",n2);}+ ?' W) ]3 D3 X! q5 F3 f% M
, s ?' s+ }- v6 p3 f+ T ) e; k5 I5 v; O$ f+ V4 g& V! m输出结果 9 K0 t2 u1 X4 X K0 B: P$ s3 ?$ R9 N* E# {
The area is 0.000000 to 1.290000.& l" l- w6 f3 R% G; N/ h
2 Q: i/ |/ b5 g5 ]# C4 k+ fThe number 1 is 8.8 T( E3 A. m4 W! _% c! r
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The min is 0.845896 and the result is -0.491089. 1 M9 m `1 K R0 [5 M& i6 f8 D2 ^: H& ?- F$ V3 J8 P- s4 H4 d+ A
The number 2 is 155 Z" y7 q$ s" j# g* s- _' C% C, f