begin
b1:=strtofloat(Edit1.text);
Q:=strtofloat(Edit5.text);
m1:=strtofloat(Edit2.text);
n1:=strtofloat(Edit4.text);
i:=strtofloat(Edit3.text);
b2:=strtofloat(Edit7.text);
b3:=strtofloat(Edit8.text);
m2:=strtofloat(Edit9.text);
m3:=strtofloat(Edit10.text);
h1:=strtofloat(Edit11.text);
n2:=strtofloat(Edit12.text);
n3:=strtofloat(Edit14.text);
for h:=0 to 1000 do
begin
W1:=(b1+m1*h1)* h1+(2*m1*h1+b1)*(h-h1);
W2:=(b2+1/2*m2*(h-h1))*(h-h1);
W3:=(b3+1/2*m3*(h-h1))*(h-h1);
x1:=2*power(sqr(h1)+sqr(m1*h1),1/2)+b1;
x2:=b2+(h-h1)*sqrt(1+sqr(m2));
x3:=b3+(h-h1)*sqrt(1+sqr(m3));
r1:=W1/x1;
r2:=W2/x2;
r3:=W3/x3;
c1:=power(r1,1/6)/n1;
c2:=power(r2,1/6)/n2;
c3:=power(r3,1/6)/n3;
Q1:=c1*sqrt(r1*i)*w1;
Q2:=c2*sqrt(r2*i)*w2;
Q3:=c3*sqrt(r3*i)*w3;
Qs:=Q1+Q2+Q3;
if
abs(Q-Qs)<eps then
Edit6.text:=FormatFloat('0',h);求h的值 ,其它都为已知,我的思路是用h试算一个Qs,当Q-Qs达到一个精度输出h,到底是用穷举法,还是试算法,我是菜鸟,希望高手指点一下!
b1:=strtofloat(Edit1.text);
Q:=strtofloat(Edit5.text);
m1:=strtofloat(Edit2.text);
n1:=strtofloat(Edit4.text);
i:=strtofloat(Edit3.text);
b2:=strtofloat(Edit7.text);
b3:=strtofloat(Edit8.text);
m2:=strtofloat(Edit9.text);
m3:=strtofloat(Edit10.text);
h1:=strtofloat(Edit11.text);
n2:=strtofloat(Edit12.text);
n3:=strtofloat(Edit14.text);
for h:=0 to 1000 do
begin
W1:=(b1+m1*h1)* h1+(2*m1*h1+b1)*(h-h1);
W2:=(b2+1/2*m2*(h-h1))*(h-h1);
W3:=(b3+1/2*m3*(h-h1))*(h-h1);
x1:=2*power(sqr(h1)+sqr(m1*h1),1/2)+b1;
x2:=b2+(h-h1)*sqrt(1+sqr(m2));
x3:=b3+(h-h1)*sqrt(1+sqr(m3));
r1:=W1/x1;
r2:=W2/x2;
r3:=W3/x3;
c1:=power(r1,1/6)/n1;
c2:=power(r2,1/6)/n2;
c3:=power(r3,1/6)/n3;
Q1:=c1*sqrt(r1*i)*w1;
Q2:=c2*sqrt(r2*i)*w2;
Q3:=c3*sqrt(r3*i)*w3;
Qs:=Q1+Q2+Q3;
if
abs(Q-Qs)<eps then
Edit6.text:=FormatFloat('0',h);求h的值 ,其它都为已知,我的思路是用h试算一个Qs,当Q-Qs达到一个精度输出h,到底是用穷举法,还是试算法,我是菜鸟,希望高手指点一下!
解决方案 »
免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货