for(i=0;fread(&eq[i],sizeof(struct eq),1,fp)!=0;i++)求解释啊.for(i=0;fread(&eq[i],sizeof(struct eq),1,fp)!=0;i++){if(strcmp(eqnum,eq[i].eqnum)!=0){fwrite(&eq[i],sizeof(struct eq),1,fp2);}}
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for(i=0;fread(&eq[i],sizeof(structeq),1,fp)!=0;i++)求解释啊.for(i=0;fread(&eq[i],sizeof(structeq),1,fp)!
for(i=0;fread(&eq[i],sizeof(struct eq),1,fp)!=0;i++)求解释啊.for(i=0;fread(&eq[i],sizeof(struct eq),1,fp)!=0;i++){if(strcmp(eqnum,eq[i].eqnum)!=0){fwrite(&eq[i],sizeof(struct eq),1,fp2);}}
for(i=0;fread(&eq[i],sizeof(struct eq),1,fp)!=0;i++)求解释啊.
for(i=0;fread(&eq[i],sizeof(struct eq),1,fp)!=0;i++)
{
if(strcmp(eqnum,eq[i].eqnum)!=0)
{
fwrite(&eq[i],sizeof(struct eq),1,fp2);
}
}
for(i=0;fread(&eq[i],sizeof(struct eq),1,fp)!=0;i++)求解释啊.for(i=0;fread(&eq[i],sizeof(struct eq),1,fp)!=0;i++){if(strcmp(eqnum,eq[i].eqnum)!=0){fwrite(&eq[i],sizeof(struct eq),1,fp2);}}
把struct的定义贴出来,另外出现什么问题了?具体
for(i=0;fread(&eq[i],sizeof(struct eq),1,fp)!=0;i++)求解释啊.for(i=0;fread(&eq[i],sizeof(struct eq),1,fp)!=0;i++){if(strcmp(eqnum,eq[i].eqnum)!=0){fwrite(&eq[i],sizeof(struct eq),1,fp2);}}
for(i=0;fread(&pil[i],sizeof(struct pilot),1,fp)!=0;i++)求整句话的示意,作用和功能!&pil[i] 的意义fp 的意思sizeof(struct pilot) 的含义
英语翻译@for(set2(i):@for(set2(j)|i#lt#j:d(i,j)*(z(i,j)+z(j,i))=@if(d(i,j)#eq#0,0,w(j))));怎么翻译成约束条件
for (n=0;!feof(fp) && fread(&tongxun[n],sizeof(struct tongxunlu),1,fp);n++);是什么意思?
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
if(fread(&s[i],sizeof(struct stu),1,fp)==1) if(fread(&s[i],sizeof(struct stu),1,fp)==1);else{printf(读取文件出错);fclose(fp);exit(0);}fclose(fp);return;调试时出现读取文件错误时为什么
分段函数 lingo@for(point(q) |q#ne#6#and#x(k,q)#eq#1#and#x(q,t)#eq#1:S=@if(m(q)#ge#0,S+m(q),S-m(q)));和@for(point(i) | x(k,i)#eq#1#and#x(i,t)#eq#1:S=@if(m(i)#ge#0#and#(S+m(i))#le#60,S+m(i),@if(m(i)#ge#0#and#(S+m(i))#gt#60,60,@if(m(i)#lt#0#and#(S+
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(
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