SETS:COORD/1..6/:X,Y,D;TRANS1/1..6/:T1;TRANS2/1..6/:T2;P1:X1,Y1;P2:X2,Y2;ENDSETSDATA:X Y D=1.25 1.25 38.75 0.75 50.5 4.75 45.75 5 73 6.5 67.25 7.75 11;ENDDATAINIT:X1 Y1=5,1;Y2 Y2=2,7;ENDINITMIN=@SUM(COORD(I):((X(I)-X1)^2+(Y(I)-y1)^2)^0.5*T1(I)+((X(I)

SETS:COORD/1..6/:X,Y,D;TRANS1/1..6/:T1;TRANS2/1..6/:T2;P1:X1,Y1;P2:X2,Y2;ENDSETSDATA:X Y D=1.25 1.25 38.75 0.75 50.5 4.75 45.75 5 73 6.5 67.25 7.75 11;ENDDATAINIT:X1 Y1=5,1;Y2 Y2=2,7;ENDINITMIN=@SUM(COORD(I):((X(I)-X1)^2+(Y(I)-y1)^2)^0.5*T1(I)+((X(I) C语言光标移动HANDLE hout;COORD coord;coord.X=3;coord.Y=3;hout=GetStdHandle(STD_OUTPUT_HANDLE);SetConsoleCursorPosition(hout,coord);重新定义X,Y 就能把光标移动到预定位置,代码看不懂, 正确的应该怎么编,还有具体的解释,#includeclass Coord {public:void setCoord(int a,int b){ x=a; y=b; }int getx(){ return x; }int gety(){ return y; }private:int x,y;};void main(){ Coord op1;int i,j;op1.x=3;op1.y=4;op1.setCoord(5,6); // 调 lingo sets中可以由已知数 和 未知数组成吗?例如 sets zuobio/1..22/:x,y; endsets 前20个已知,怎么写 lingo 关于and的用法sets:jishu/1..n/:x,h,y;endsets当i>1且i model:title Location problem; sets:demand/1..6/:a,b,d; supply/1,2/:x,y,e; link(demand,supply):c; e照书上写的,还是出错了.大惑不解.错误代码：50for函数使用不当所用版本为lingo11model:title Location problem;sets:demand/1..6/: LINGO分段函数怎么求!sets:position/1..7/:x,f;society/1..15/:r,Y;endsetsdata:r=2 14 3 6 9 4 8 12 10 11 6 14 9 3 6;f=9 6.5 20 14.5 19 13 10.5;enddatamax=@sum(society(i):y(i));Y(1)=@if(x(1)#eq#1,0.7*r(1),0);Y(2)=@if((x(1)+x(2))#le#0,0,@if((x(1)+x( 帮忙看下这个lingo程序哪里出错了MODEL:SETS:CUTFA/1..6/:x;CUTFAA/1..11/:y;切割的方法有17种,x表示对应1-6种切割方法的原料钢管的需求量,y表示对应7-17种切割方法的原料钢管的需求量;BUJ/1..3/:n,b;3种钢 关于c++的问题,#include void gotoxy(int x,int y){COORD pos = {x,y};HANDLE hOut = GetStdHandle(STD_OUTPUT_HANDLE);SetConsoleCursorPosition(hOut,pos);}enum {Up = 72,Down = 80,Left = 75,Right = 77,Space = 32 };我知道这个函数作用是把光标 lingo错误代码11：a syntax error has occuredMODEL:Title Location Problem;sets:demand/1..6/:a,b,d;supply/1,2/:x,y,e;link(demand,supply):c;endsetsdata:a=1.25,8.75,0.5,5.75,3,7.25;b=1.25,0.75,4.75,5,6.5,7.75;d=3,5,4,7,6,11;e=20,20;enddatainit:x,y=5, 帮我看看lingo程序有没问题.说是@for语句不当model:sets:liaochang/1,2/:x,y,e;gongdi/1..6/:a,b,d;links(liaochang,gongdi):c;endsetsmin= @sum(links:c(i,j)*((x(i)-a(j))^2+(y(i)-b(j))^2);@for(gongdi(j):@sum(liaochang(i):c(i,j))=d(j));@for(lia lingo程序错误1017model:sets:question(/1..i/):x,y;endsetsinit:x(1)=0;y(1)=0;endinit[obj] max=@sum(link(i,j):c(i))@for(question:(min(@sqrt((x(i)-x(i1))^2+(y(i)-y(j))^2))>=z););@for(question:(max(@sqrt((x(i)-x(i1))^2+(y(i)-y(j))^2)) 看看这个lingo程序哪里出错了?model:sets:de/1..8/:x,y;fe/1..5/:f;link(fe,de):W;endsetsdata:x=5 7 5 4 6 5 5 3;y=25 36 32 15 31 28 22 12;L=20;enddatamin=@sum(de:x*w(1,j))+@sum(de:x*w(2,j))+@sum(de:x*w(3,j))+@sum(de:x*w(4,j))+@sum(de:x*w(5,j))+ WGS84（中央子午线111度,3度带37带） 纬度　(D:M:S):39:48:29.47097N;经度　(D:M:S):110:03:57.05204E80坐标（中央子午线111度,3度带37带） X:4408638.91 Y:419912.773用COORD坐标转换,只设置中央子午线为111,在主 lingo中不是可以取消变量的非负限制吗,为什么老是出错啊Model:Title Location Problem;sets:demand/1..6/:a,b,d;supply/1..2/:x,y,e;link(demand,supply):c;endsetsdata:a=1.25,8.75,0.5,5.75,3,7.25;b=1.25,0.75,4.75,5,6.5,7.75;d=3,5,4,7,6,1 lingo中不是可以取消变量的非负限制吗,为什么老是出错啊,Model:Title Location Problem;sets:demand/1..6/:a,b,d;supply/1..2/:x,y,e;link(demand,supply):c;endsetsdata:a=1.25,8.75,0.5,5.75,3,7.25;b=1.25,0.75,4.75,5,6.5,7.75;d=3,5,4,7,6, 帮忙看下一个Lingo程序,自己编的运行不了了具体如下：MODEL:SETS:JOB/1..5/;WORKER/1..7/;REVISOR/1..7/;LINKS(WORKER,REVISOR,JOBA,JOBB,TIME,ZZ):C,D,X,Y,T,Z;ENDSETSDATA:C=2 15 13 1 810 4 14 15 79 14 16 13 87 8 11 9 48 4 15 8 612 4 6 8 lingo程序：model：sets：product/1..3/:a,b; row/1..5/:c,d,y; num/1..9/:x; endsets其中的product和num该怎么用?还有平时不是用的col么而不是num区别在哪?