网友您好, 请在下方输入框内输入要搜索的题目:

题目内容 (请给出正确答案)
I need ( ) cloth, for I'm going to make ( ) clothes.

A、many/ many

B、many/ a lot o{

C、a lot of/ many

D、much/ much


参考答案

更多 “ I need ( ) cloth, for I'm going to make ( ) clothes. A、many/ manyB、many/ a lot o{C、a lot of/ manyD、much/ much ” 相关考题
考题 ---- Shall we stay at home watching TV tonight?---- No, I’d like __________ and see a movie.A. to go B. go C. went D. going

考题 —_________ I take the book out?—I'm afraid not.A. WillB. MayC. MustD. Need

考题 –Do you ever play tennis?–_______.I have played a few times.I'd rather _____ jogging. A.Hardly ever ⋯goingB.Hardly ever ⋯goC.Hardly never ⋯goD.Hardly never ⋯going

考题 It was raining outside, I ( ) one umbrella and walked out.A.picked upB.take upC.haveD.make

考题 I ________ help. I can do it myself. A.need toB.don’t needC.needn’tD.need

考题 I ____ you will write me back soon. A、wishB、hopeC、wantD、need

考题 在Need矩阵中,need(i,j)=a表示()。A.进程i最多需要j类资源a个B.进程i最多还需要j类资源a个C.进程j最多还需要i类资源a个D.进程j最多需要i类资源a个

考题 2、在Need矩阵中,need(i,j)=a表示()。A.进程i最多需要j类资源a个B.进程i最多还需要j类资源a个C.进程j最多还需要i类资源a个D.进程j最多需要i类资源a个

考题 下面()仅输出m的大于1的最小因子。A.for (i =2; i<=m-1; i++) if (m % i == 0) { printf("%d is 最小因子n", i); break; }B.for (i =2; i<=m-1; i++) if (m % i == 0) { printf("%d is 最小因子n", i); continue; }C.for (i =2; i<=m-1; i++) if (m % i == 0) { printf("%d is 最小因子n", i); }D.i=2; while (m % i != 0) i++; printf("%d is 最小因子n", i);

考题 下面()是正确的判断素数程序(m>1)。A.j=0; for (i =2; i<=m-1; i++) if (m % i != 0) j++; if(j==m-2) printf(“%d是素数n", m);B.j=0; for (i =2; i<=m-1; i++) if (m % i == 0) j++; if(j==0) printf(“%d是素数n", m);C.flag=0; for (i =2; i<=m-1; i++) if (m % i == 0) flag=1; if(flag==0) printf(“%d是素数n", m);D.for (i =2; i<=m-1; i++) if (m % i == 0) i=m+2; if(i==m+3) printf(“%d是素数n", m);