I want to learn() about the industry so that I'm better prepared. A、possible ofB、as much as possibleC、as possible muchD、as possible as much
I want to learn() about the industry so that I'm better prepared.
A、possible of
B、as much as possible
C、as possible much
D、as possible as much
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听力原文：M: So, Jane, how long have you been an author?W: Well, Tom, I didn't start writing until I was in my thirtieth, and I'm over seventy now. So goodness, I must have been writing for about forty years.How long has the woman been an author?A．About 30 years.B．About 40 years.C．About 60 years.D．About 70 years.
下面程序输出的结果是()。includeusing namespace std;void main(){ char ch[][8]={"g 下面程序输出的结果是( )。 #include＜iostream＞ using namespace std; void main() { char ch[][8]={"good","better","best"}; for(int i=1;i＜3;++i) { cout＜＜ch[i]＜＜endl； } }A．good betterB．better bestC．good bestD．good
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);
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);