lingo @for(var(i):k(i)=@sum(var_1:c*w));有什么语法错误

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lingo @for(var(i):k(i)=@sum(var_1:c*w));有什么语法错误
lingo @for(var(i):k(i)=@sum(var_1:c*w));有什么语法错误

lingo @for(var(i):k(i)=@sum(var_1:c*w));有什么语法错误
@for(var(i):k(i)==@sum(var_1:c*w));

lingo @for(var(i):k(i)=@sum(var_1:c*w));有什么语法错误 lingo这个错误报告index variables may not share names with other variables是怎么回事?model:sets:var/1..12/:a;obj(var,var):x;endsetsdata:a=2.8 3.2 1.2 4 2.70.9 2.5 3.3 1.7 2.5 3.6 4.7;enddatamin=@sum(var(i):@sum(obj(k,l)|k#le#i:x(k,j)*a(j))) for(var i=0;i for循环的疑问function myFunction(){var x=;for (var i=0;i 英语翻译var i,j;for(i=1;i lingo中@for函数用错@for(team(k):@sum(link(i,j):x(i,j,k))=10;); 请问这一句错了吗?错哪 function searchType( type ){for(var i=1;i Math.random是什么意思for (var i:Number=0; i program Noname0; const n=7; var a:array[1..n] of integer; i,j,k,t:integer; begin write('ggg'); for program Noname0;const n=7;var a:array[1..n] of integer;i,j,k,t:integer;beginwrite('ggg');for i:=1 to n do read(a[i]);for i:=1 to n-1 dofor j:=n downto 、这段lingo代码什么意思 @for(points(p):@sum(points(i)|i#ne#p:@sum(cars(k):x(i,p,k)))=1;@sum(points(j)|j#ne#p:@sum(cars(k):x(p,j,k)))=1;); 关于 LINGO @FOR 语句请问下面的@for 语句中为什么@for(set1(i):@sum(set1(k):x(k))=1); i j k 总是在换啊?model:sets:set1/1..2/:x;set2/1..2/:y;link(set1,set2):a,b;endsets@for(set1(i):@sum(set1(k):x(k))=1);@for(set2(j):@sum(set2(k):y(k)) lingo 中如何表示一个矩阵X(k,i,j) lingo中if语句@if(@abs(z(i)-k(j)) 为什么这一步lingo提出我missing right parenthesis的错误 实在看不出 @for(trips(j):@for(trips(i)|i#ne#j:@for(camps(l):((beginingmoment(j)+l*periodtime(j)+@sum(camps(k)|k#lt#l:time(j,k))>=beginingmoment(i)+l*periodtime(i)+@sum(camps(k)|k lingo语句哪错了sets:position/P1 P2 P3 P4 P5/:x,k;endsetsdatax=5 6 4 8 10;enddata@for(position(i):k(i)=1/x(i)); 十进制实数转二进制的程序Program zy6;var bin:array [0..50] of 0..1;x:longint;k,i:integer;beginwriteln('read decimal data x:'); read(x);for i:=0 to 50 do bin[i]:=0;k:=0;while x>0 dobeginbin[k]:=x mod 2;x:=x div 2;k:=k+1;end;writeln('Binary: 这段lingo程序的意思?@sum(point_aim(i):q(i)*@sum(cars(k)|k#eq#1:y1(i,k))) lingo利用for循环,有错误1 lingo有没有真为1这么一说2 @for(lianxi(i,j)| i #NE# j:(T(i)#EQ#1)*@floor((i+4)/5) #NE# (T(j)#EQ#1)*@floor((j+4)/5)); (j+4)/5))); 这句话哪有错误.整个程序是model:sets:jiaoshi/1..45/:dianliang,zuowei,K