lingo程序运行错误model:sets:bed/bed1..bed7/;lebe/leb1..leb3/;opera

程序.C(174):warning C209:'_display':too few actual parameters
void main()
{
lcd_init();
measure();
display();
aa=1;
bb=1;
ff=2000;
while(1)
{
ff=fre;
if(a[4]!=0)
{
ff=(a[0]+a[1]+a[2]+a[3]+a[4])/5;
}
//display(l,0,ff);if(TFl==l)
//当频率远超过测量范围计数溢出时（TF为计数器1溢出位）,电容值为0
{
c=0;
}
else if(TF1==0)
{
if(ff10000)
if(aa==0&&bb==0)
{ r1=480;
aa=bb=1;
}
js();
if(c1300||c3000||c2000||c

c程序 abs函数
//main.c
#include
void main()
{
double i=abs(2.5-2.7); //结果i=0.00000000000000000,为什么?
}

java 程序阅读 Parent
class Parent
{
Parent(int j)
{
}
}
class Child extends Parent
{
Child(int i)
{
}
public static void main(String args[])
{
Child c=new Child(3);
}
}
这段代码错在哪里?

程序哪里错误



model:
sets:
j/1..5/:danjia,yongliao,chengben,xuqiu,shelitouzi,fj;
i/1..5/:chaixienengli,danweichaixiehuafei,chuzhihuafei,fenjieziyuanshiyong,vi,ti;
k/1..5/:uk,vk,bk;
l/1..5/;
links1(i,l):oil,pil,gil,lzuidanengli,xil,uil;
links2(i,j):qij;
links3(i,k):rik,bik,sik,qik;
endsets

解读VFP 程序replace all 数量 with 数量 -b->数量
select 2
use sale
index on 商品名 tag 商品名
select 1
use ing
set relation to 商品名 into sale
replace all 数量 with 数量 -b->数量,总金额 with 单价*数量
use in 1
use in 2
里面的
-b-> 什么意思.请详细介绍谢谢.
replace all 数量 with 数量 -b->数量,总金额 with 单价*数量
那这句话又是怎么翻译的谢谢啦

如何使用循环语句缩短程序
subplot(521);
plot(index,a1);
subplot(522);
plot(index,a2);
subplot(523);
plot(index,a3);
subplot(524);
plot(index,a4);
subplot(525);
plot(index,a5);
subplot(526);
plot(index,d1);
subplot(527);
plot(index,d2);
subplot(528);
plot(index,d3);
subplot(529);
plot(index,d4);
subplot(5,2,10);
plot(index,d5); 以上是程序的一部分,我想问下可不可以把这段程序用循环语句简练的表达出来,这样实在是太麻烦了,麻烦大神老师们给看看,跪谢!

跪求高手帮忙解答ACM程序代码
The input file will contain one or more lines, each line containing one integer n with , representing the number of cities in the country.
Input is terminated by a value of zero (0) for n.
输入 3 4 5 6 7 8 9 10 11 12 0 输出 2 5 2 4 3 11 2 3 8 16
To put an end to this problem, the University of Ulm has developed a contingency scheme for times of peak load to cut off net access for some cities of the country in a systematic, totally fair manner. Germany's cities were enumerated randomly from 1 to n. Freiburg was number 1, Ulm was number 2, Karlsruhe was number 3, and so on in a purely random order.
Then a number m would be picked at random, and Internet access would first be cut off in city 1 (clearly the fairest starting point) and then in every mth city after that, wrapping around to 1 after n, and ignoring cities already cut off. For example, if n=17 and m=5, net access would be cut off to the cities in the order [1,6,11,16,5,12,2,9,17,10,4,15,14,3,8,13,7]. The problem is that it is clearly fairest to cut off Ulm last (after all, this is where the best programmers come from), so for a given n, the random number m needs to be carefully chosen so that city 2 is the last city selected.

lingo 程序出错 syntax error has occurred
max=4820*(1-0.1*4*a1)*297*(1+a1)*1000+320*(1-0.1*27*a2)*720*(1+a2)*1000+210(1-0.1*11*a3)(1+0.1*a4)*(1+a3)*1050*1000+70*(1-0.1*4*a4)(1+0.1*4*a3)*815*(1+a4)*1000
4820*(1-0.1*4*a1)*297*0.01*4+320*(1-0.1*27*a2)*0.01*80+210(1-0.1*11*a3)(1+0.1*a4)*0.01*35+70*(1-0.1*4*a4)(1+0.1*4*a3)*0.01*25

运行程序lingo程序,出现unexpected jacobian overflow
min=0.125*@sqrt(x1^2+(x2-x1)^2)+0.125*@sqrt((x2-x1)^2+(x3-x2)^2)+0.125*@sqrt((x4-x2)^2+x3^2)+0.125*@sqrt(x1^2+x4^2)
+0.125*@sqrt((x1-5)^2+5^2)+0.125*@sqrt((x1-5)^2+(x2-5)^2)+0.125*@sqrt((x2-5)^2+(x3-5)^2)+0.125*@sqrt((x4-5)^2+5^2)
x1

C++ 程序设计习题 验证极限,
关于输入
三个数,分别代表x,a和e,其中a>1,e>0
关于输出
输出最小的非负整数N,使得当整数n>N时,有|x/a^n| < e成立
我的程序是：
#include
using namespace std;
int main()
{
double a,e,x,y;
int N;
cin>>x>>a>>e;
for (N=0;;N++)
{
y=pow(a,N);
if (abs(x/y)

运行lingo程序出现unexpected jacobian
model:
min=max(T1..T5);
n1*8+2*((k1-1)*3.2+4.8)/1.6*x1=T1;
n2*8+2*((k2-1)*3.2+4.8)/1.6*x2=T2;
n3*8+2*((k3-1)*3.2+4.8)/1.6*x3=T3;
n4*8+2*((k4-1)*3.2+4.8)/1.6*x4=T4;
n5*8+2*((k5-1)*3.2+4.8)/1.6*x5=T5;
k5=28;
@bnd(1,k1,k2);
@bnd(k1,k2,k3);
@bnd(k2,k3,k4);
@bnd(k3,k4,k5);
@gin(k1);@gin(k2);@gin(k3);@gin(k4);
n1=0.5*(k1-1);
n2=0.5*(k2-k1);
n3=0.5*(k3-k2);
n4=0.5*(k4-k3);
n5=0.5*(k5-k4);
x1=(k1-1)*100/18;
x2=(k2-k1)*100/18;
x3=(k3-k2)*100/18;
x4=(k4-k3)*100/18;
x5=(k5-k4)*100/18;
end

C+ +程序,限制输入长度,保留小数位数
1.设圆半径r=1.5,圆柱高h=3,圆柱体积v
用cin输入数据,输出计算结果,输出时要求有文字说明,输出长度为setw（10）,取小数点后2位数字,右对齐.

lingo 程序
sets:
k/1..5/;
z/1..8/:m;
link(k,z):a,c,l,t,s;
endsets
min=a(i,j);
@sum(z(j):a(4,j)*c(4,j))

写出下列程序运行结果. LET A= -5 LET B= 4 LET C=ABS(A)+B PRINT C END 运行结果:_______

判断奇偶程序
#include
int main()
{
int a,b;
printf("输入a=");
scanf("%f",&a);
b=a%2;
if(b==0)
printf("偶n");
else
printf("奇n");
}
运行程序出错,错在哪里?为什么把%f改成%d就对了?

matlab遗传算法程序解读?
%GA(Generic Algorithm) Program to optimize PID Parameters
clear all;
close all;
global rin yout timef
Size=30;
CodeL=3;
MinX(1)=zeros(1);
MaxX(1)=20*ones(1);
MinX(2)=zeros(1);
MaxX(2)=1.0*ones(1);
MinX(3)=zeros(1);
MaxX(3)=1.0*ones(1);
Kpid(:,1)=MinX(1)+(MaxX(1)-MinX(1))*rand(Size,1);
Kpid(:,2)=MinX(2)+(MaxX(2)-MinX(2))*rand(Size,1);
Kpid(:,3)=MinX(3)+(MaxX(3)-MinX(3))*rand(Size,1);
G=100;
BsJ=0;
%
Pm=0.10-[1:1:Size]*(0.01)/Size; %Bigger fi,smaller Pm
Pm_rand=rand(Size,CodeL);
Mean=(MaxX + MinX)/2;
Dif=(MaxX-MinX);
for i=1:1:Size
for j=1:1:CodeL
if Pm(i)>Pm_rand(i,j) %Mutation Condition
TempE(i,j)=Mean(j)+Dif(j)*(rand-0.5);
end
end
end
%Guarantee TempE(Size,:) belong to the best individual
TempE(Size,:)=BestS;
Kpid=TempE;
end
Bestfi
BestS
Best_J=BestJ(G)
figure(1);
plot(time,BestJ);
xlabel('Times');ylabel('Best J');
figure(2);
plot(timef,rin,'r',timef,yout,'b');
xlabel('Time(s)');ylabel('rin,yout');

1²+2²+.+99²+100²帮忙设计算法程序框图
如果不能画出框图的话 请用算法的步骤解答（即第一步……,第二步……,……如此类推）谢谢

这个程序为何无法交换A,
#include"stdio.h"
int a,b;
int ch(int a,int b)
{ int t;
t=a;
a=b;
b=t;
return(a,b);
}
int main()
{
scanf("%d,%d",&a,&b);
ch(a,b);
printf("%d,%d",a,b);
}
不使用指针,
利用全局变量不是可以达到交换的吗?为何这个程序达到不了交换的目的呢?

菜鸟程序改错
# include
# include
int main(void)
{ int a = 1;
int b = 2;
int c = 1;
double delta;
double x1 ;
double x2 ;
int delta = b*b-4*a*c;
if ( delta > 0 )
{ x1= (-b + sqrt(delta)/(2*a);
x2= (-b - sqrt(delta)/(2*a);
printf ("该一元二次方程有两个解,x1 = %f,x2 = %fn",x1,x2);
}
else if ( delta == 0 )
x05{ x1=(-b)/(2*a);
x2 = x1 ;
x05x05printf ("该一元二次方程有唯一解,x1 = x2 = %fn",x1);
x05x05
x05x05}
else
x05{ printf ("无解"n",x1); }
x05return 0;
}
-------------------Configuration:Cpp2 - Win32 Debug--------------------
Compiling...
Cpp2.cpp
C:Documents and SettingsAdministrator桌面新建文件夹Cpp2.cpp(12) :error C2371:'delta' :redefinition; different basic types
C:Documents and SettingsAdministrator桌面新建文件夹Cpp2.cpp(9) :see declaration of 'delta'
C:Documents and SettingsAdministrator桌面新建文件夹Cpp2.cpp(15) :error C2143:syntax error :missing ')' before ';'
C:Documents and SettingsAdministrator桌面新建文件夹Cpp2.cpp(16) :error C2143:syntax error :missing ')' before ';'
C:Documents and SettingsAdministrator桌面新建文件夹Cpp2.cpp(26) :error C2017:illegal escape sequence
C:Documents and SettingsAdministrator桌面新建文件夹Cpp2.cpp(26) :error C2001:newline in constant
C:Documents and SettingsAdministrator桌面新建文件夹Cpp2.cpp(26) :error C2146:syntax error :missing ')' before identifier 'n'
C:Documents and SettingsAdministrator桌面新建文件夹Cpp2.cpp(30) :fatal error C1004:unexpected end of file found
执行 cl.exe 时出错.
Cpp2.obj - 1 error(s),0 warning(s)

Visual Basic程序问题
与程序段：
a = a + b
b = a - b
a = a - b
功能相同的程序段是
A．a = b
b = a
B．a = a + b
a = a - b
b = a - b
C．t = a
a = b
b = t
D．If a > b Then
a = b
Else
b = a
End If
答案是C,为什么呢?

lingo高手帮忙,程序出现Unexpected jacobian overflow错误
model:
sets:
time/1..7/:t,w;
power/1..4/:pmax,pmin,TC,C,MC,TN;
link(time,power):p,A,B,n;
endsets
data:
t=6,3,3,2,4,4,2;
w=10000,30000,22000,33000,22000,27000,16000;
pmin=750,1000,1200,1800;
pmax=1750,1500,2000,3500;
C=2250,1800,3750,4800;
TC=4000,1500,2200,1300;
MC=2.6,2.1,1.7,3.6;
TN=9,4,7,3;
enddata
min = @sum(links(i,j):(n(i,j)*t(i)*C(j)+t(i)*MC(j)*(p(i,j)-pmin(j))+A(i,j)*TC(j));
@for(time(i):@sum(power(j):(n(i,j)*p(i,j)=w(i))));
@for(power(j):@bnd(time(i):pmin(j),p(i,j),pmax(j));
@for(links(i,j):@gin(n(i,j)));
@for(power(j):time(i):n(i,j)

Mathematica 作图程序
留的数学期末试验.不过我们根本没学过 Mathematica,
作图 y=sinX y=cosX
极限 lim1/X(x趋近于无穷大） limsinX（x趋近于0）
微分 对y=sinX 和 y=1/X求微分
积分 求y=cosX的原函数 不定积分3x
我就40 分了
Mathematica程序：
运行结果：

lingo程序纠错 错误代码63
sets:
time/1..3/;
document/1..3/;
link(time,document):a;
endsets
min=z;
z>=a(t,j)*t;
@gin(t);
@gin(j);
@for(document(y):
@sum(time(x):a(x,y))=1);
a(t,1)+a(t,2)+a(t,3)<=1;

lingo程序问题
@for(areas(i):
@for(schools(j):
as=@sum(expenses(i,j):m+n+k)));
@for(areas(i):
@for(schools(j):
as(i,j)#ge#50 #or# as(i,j)#eq#0 =1));
as代表各地区到各学校上课的人数,我这里想约束as要么大于或等于50,要么等于0.但一直有问题
是没有可行解,但是我去掉这一约束,将不符合约束的一个个限制,就可以运行成功

QBASIC试题~
写出下列程序运行结果.
LET A=1
LET B=5
LET T=A
LET A=B
LET B=T
PRINT B
END
运行结果:(_1_)