putpixel(i,j,1);i_val=655___35;local->tm_min,local->tm_sec,RMess);ctstudent*head,intag

putpixel(x j,y i,pixel_save[i][j]);MouseObig_mem[i]=big_mem[i 1];==0 putpixel(i,j,1);i_val=655___35;local->tm_min,local->tm_sec,RMess);ctstudent*head,intag if(*out==*(out 1))putpixel(i,j,1);main()longge,shi,qian,wan,x;ACTIVE_DATABAS 想知道：putpixel(x j,y i,0);bar(226,231,512,330);structout_product*p;序上了,开始工作 跪求cpuPoker[index][j];putpixel(MouseX,MouseY,YELLOW);ch=str[0];floatpeven(intn)floats; for(j=i+1;j 怎么用lingo求解多目标规划呢,max =@sum(I(i):@sum(J(j):@sum(link(I,J):x(i)*w(j)*p(i,j)/d(i,j))));@for(I(i):@sum(J(j)|w(j)#eq#1:x(i)*p(i,j))>0);@for(I(i):@sum(J(j):x(i)*p(i,j)*c(j))>0);@for(J(j):@sum(link(I,J):x(i)*p(i,j)>=0));@for(l int i=2,j=1,k=3 i&&(i+j)&k|i+j string.Format({0}*{1}={2},i,j,i*j); if语句用find函数怎么优化,程序如下：for i = 1:mfor j = 1:nif abs(R(i,j)-R1(i,j))>0.05R2(i,j) = R1(i,j)elseR2(i,j) = 0endif abs(G(i,j)-G1(i,j))>0.05G2(i,j) = G1(i,j)elseG2(i,j) = 0endif abs(B(i,j)-B1(i,j))>0.05B2(i,j) = B1(i,j)elseB2(i,j) int i,j=2;*p=&i;*p=(*&j+1)^j; i=? 阻抗 j的含义I=-j/1-j 中的j for( int i=0,j=1; j < 5; j+=3 ) i=i+j; 为什么等于5 for(i=1;i<10;i++){for(j=1;j<=i;j++)解释一下这是什么意思? main() {int i,j,w=0; for(i=4; i; i--) for(j=1;j lingo程序求解释~n=@size(cities); min=@sum(link(i,j)|i#ne#j :distance(i,j)*x(i,j)); @sum(cities(i)|i#gt#1 :x(1,i))>=1; @for(cities(i)|i#gt#1:@sum(cities(j)|j#ne#i:x(j,i))=1; @for(cities(j)|j#gt#1#and#j#ne#i:level(j)>=level(i)+x(i,j)-(n-2)*(1-x(i, matlab多元函数求最小值问题matlab求r =1/(9-j)*(i^2+81-18*j+j^2)^(1/2)*(38/5-19/20*fix(j)+1/20*(8-fix(j))^2)+(fix(j)+1-j)/(9-j)*(i^2+81-18*j+j^2)^(1/2)*(9/5-1/10*fix(j))+(j^2-2*j*n+n^2+m^2-2*m*i+i^2)^(1/2)/(j-n)*(-19/20*fix((n*i-m*j)/(i+n-j- xx[i][j]=xx[i][j]>>4+xx[i][j-1];和 xx[i][j]=（xx[i][j]>>4）+xx[i][j-1];的结果为什么不一样啊