6 ?" O C2 v- u% L. ^9 g4 l1 l% @void main() 3 O! G/ Z* n- ]& ^3 H- h! v7 o( |$ {- G8 n, U
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float xx, ff; 1 J# o: ]" [! a! P5 k( b) x5 u 9 s j. ~6 [( c: Fxx=gold(&ff); 6 `+ S6 _8 ~' l s" [( q8 V) K( t2 @; r) K/ q
printf("\nThe Optimal Design Result Is: \n"); % v4 z$ b# B0 f% Q& U " ^3 x8 J9 j6 E) D ? eprintf("\n\tx*=%f\n\tf*=%f",xx,ff); 7 O0 k2 l; h( j7 n- X9 g5 a9 c; |0 p' r( s
getch();/ v: y9 I# v" s. z8 B" z& G7 o
- F$ {1 G; Y+ g; P7 M' ^} ( T9 V4 r- p$ G$ L7 z# X+ V; C # c: \; D) Y& M: C2 ^4 I8 q, ]; H #include <stdio.h> & C! r6 J; @8 s$ a- ~#include <math.h> a' U) M6 v0 K5 J #define e 0.0014 l6 L9 y5 U' W #define tt 0.001! Q* H% r5 V1 Z3 C$ J8 a7 q" [ float f(double x) * C V& f/ g" i; M$ d{* a4 {+ c i8 I X; r) T! `9 p float y=9*pow(x,4)-7*pow(x,3)-4*pow(x,2) +2;% }/ }8 N. [9 \: G: I E+ @ return(y);2 g- u4 r9 O) c$ \' L; m3 M }" W& l: N& d- b5 {! x( V finding(float *p1,float*p2), H3 u2 A* v W" L& l* Z: V {4 e% Q. Q- r/ V2 d float x1=0,x2,x3,t,f1,f2,f3,h=tt; 4 r" {. j4 \. p4 Y7 I3 o9 F# Bint n=0;: Z; X+ k) @+ Q% h7 d! p. z. k" ~ x2=x1+h;f1=f(x1);f2=f(x2);: q+ U+ `1 O) K$ h if(f2>f1) {h=-h;t=x2;x2=x1;x1=t;} & U' A6 i+ n) W; f: V; ado + x) H# }) U8 a+ V) |3 [. \, Q3 S# m( t* ^8 N9 F$ A9 E
{ x3=x2+h;h=2*h;f3=f(x3);n=n+1;} 6 M! } i2 B, Q; ?while(f3<f2);1 K1 P8 [6 N" h$ m# d5 `$ j, d if(x1>x3) {t=x1;x1=x3;x3=t;}9 I5 f. [0 @0 u# e *p1=x1;*p2=x3;7 j; c+ y1 H4 ?8 V& K return(n);- m# R0 ^- z/ m+ w" S4 C9 Y }+ O9 H3 Y, l- l& @4 q: B$ `% M M1 a gold(float *p)0 X9 s, E1 R) N. o: c. @2 ^ { & Y1 V' O4 t9 t! m1 s/ lfloat a,b,x1,x2,f1,f2;7 I+ m- Z7 j8 l; z3 `) g int n=0; 9 R% M" |" }, Q, ~2 v; Yfinding(&a,&b); e! `% D# L( L- l' f; W1 ` do/ F. G% a3 Y- V, K% m0 Q) R ( w( p+ P$ g+ {# G- |/ [5 R) K
{x1=a+0.382*(b-a);- Z( w; P" S O6 J : b! Y# [7 c* P, [x2=a+0.618*(b-a);f1=f(x1);f2=f(x2);n=n+1; ! |( [( N0 _ N , F& g! b- Q$ N2 G+ wif(f1>f2) a=x1; 0 P v# w% B# w! ~% m$ S$ a : L) W) w3 c& W0 G9 u6 yelse b=x2;} + I, c. B- Q3 ^- bwhile((b-a)>e); 4 N# N4 m' B, O$ m, y. {* F*p=(x1+x2)/2;return(n); $ ?( @- S4 ~/ \0 @9 P} 8 \$ _/ K5 n% O+ o* |# \' A/ u. N4 Z8 ?# I
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4 y! L5 }% Z3 s) ]main()& p; ^3 T* _0 ]; h$ o! P { 6 u% ]5 s2 G: s9 `' j5 @) c7 Rfloat a,b,x,min;int n1,n2; , ]: s2 E. F9 ]* f: l1 `+ [) in1=finding(&a,&b);, a1 A- V; \% F4 R9 F2 H& r n2=gold(&x); 8 E t' p) z6 S: {/ ~/ F. Rmin=f(x); ! M/ l# V2 l' E$ m+ z- v! jprintf("\n The area is %f to %f.",a,b); printf("\n The number 1 is %d.",n1);, E& O7 e6 R! o+ [7 m; }9 [9 X! M printf("\n The min is %f and the result is %f.",x,min); ( @% Q* m$ E/ x+ L, G% G4 l. y, ]7 Fprintf("\n The number 2 is %d.",n2);} , |& K: @/ I4 ^" n E- |( A: b6 t8 o: D7 o/ e8 ^% o/ y6 H
/ U' l8 D8 o0 h) e7 ~* c# y输出结果 " c; G! p# a r. m- O+ P; S) [" ]1 _. e) T* L* U0 c2 O
The area is 0.000000 to 1.290000. 7 _3 E7 k: v+ q2 q$ O r" V* @6 c4 U0 G. R' K7 Q7 g
The number 1 is 8.; @! S3 C: s- h! r+ [
% J7 b- E* p8 |0 hThe min is 0.845896 and the result is -0.491089., \& b# x6 V8 B# b8 p( k6 k, f
% \8 s/ b; _! O" _: `, NThe number 2 is 155 k! f1 s5 S* E