这段lingo程序的意思?@sum(point_aim(i):q(i)*@sum(cars(k)|k#eq#1:y1(i,k)))

这段lingo程序的意思?@sum(point_aim(i):q(i)*@sum(cars(k)|k#eq#1:y1(i,k))) 、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;); 一个lingo程序问题：s=@sum(number(I) | I #le# 5:x);表示啥意思? 请懂Lingo的人帮忙解释一下这段代码的意思,min=@sum(zijin:y*a)+@sum(yaoqiu(i):d(i)*0.5*@min(link(i,j):y(i)*k(i,j))); @for(zijin:@bin(y));sets:zijin/1..49/:y,a;yaoqiu/1..49/:d;link(build,require):k;endsetslink(zijin,yaoqiu): LINGO软件的基本语法,比如@sum, lingo的数据拟合model:sets:xl/1..5/:n,y; endsets data:n=19,20,22,24,25; y=6,7,8,9,10; enddata @frmodel:sets:xl/1..5/:n,y;endsetsdata:n=19,20,22,24,25;y=6,7,8,9,10;enddata@free(a);@free(b);min=@sum((@log(y)*a-b-n)*(@log(y)*a-b-n));end这段程序 执行下列程序段后,sum的值是?int sum; int a=10,b=9; sum=(a--==++b)?100:200; 2-17 执行下列程序段后,sum的值是 .for (i = 1; i lingo中@sum(m: 请高手帮我找下这个Lingo程序的错误!下面是Lingo程序model:sets:hotel/1..10/;format/1..3/;link(hotel,format):x,y,c;endsetsmin=@sum(link:x*y*c);@for(format(j):@sum(hotel(i):(y(i)*r(i,j)))=num(j)/2);@for(format(j):@sum(hotel(i):(y(i)*r1( 那位高手知道下面这段lingo程序为什么解不出来?谢谢,急!model:sets:warehouses/wh1..wh6/:capacity;vendors/v1..v18/:demand;links(warehouses,vendors):cost,volume,x;endsets!目标函数;min=@sum(links:cost*volume*1);!需求约束;@for(v 线性最优化的LINGO程序举例 请问这段Lingo程序错在哪里?MODEL:SETS:Add/1 2 3 4 5 6/:p;Demand/1 2 3 4 5 6/:d;Num/1 2 3 4 5 6/:x;Left/1 2 3 4 5 6 7/:y;ENDSETSDATA:p=0.1 0.1 0.2 0.2 0.3 0.3;d=42 32 41 67 25 29;ENDSETSmin=@SUM(0.5*(1+Add(I):p(I))*(Num(I):x(I))^2)+@SUM(0.05* lingo的运行错误问题我的lingo程序模型是这样：MODEL:SETS:QUARTERS/1..1000/:X;ENDSETSMIN = @SUM(QUARTERS(i)|@MOD(i,2)#EQ#0:X(i);) - @SUM(QUARTERS(i)|0#NE#@MOD(i,2):X(i););@for(QUARTERS(i):@sum(QUARTERS(i):X(i)) >= 1000 ;);@for(QUARTERS( 一下程序段运行后,变量sum的值是sum=0i=0For i=2 To 10 Step 2sum=sum+iNext i怎么算的,列数学算式给我. C语言 DO WHILE.下面程序段执行后的结果为:( ) intsum=1,n=10; do { sum=sum*n; n++; } while(n sum的意思是什么 如下程序段的运行结果,我算的是110,答案是109,为什么main（）{int i=5,sum=100;do{sum=sum+(--1);}while(i>=0);printf(%d,,i);}我改了下,如果有错,请帮忙指出,哪里错了main（）{int i=5,sum=100;do{sum=sum+(--i);}while(i>