I’m expecting a letter____Jimmy.A.forB.fromC.withD.into

题目
I’m expecting a letter____Jimmy.

A.for

B.from

C.with

D.into

相似考题

1.Project work packages are typically decomposed into smaller components called activities to provide a basis for()scheduling, executing, and monitoring and controllingthe proj ect work.A.reviewingB.estimatingC.auditingD.expecting

2.● Project work packages are typically decomposed into smaller components called activities to provide a basis for （72）, scheduling, executing, and monitoring and controlling the project work.（72）A. reviewingB. estimatingC. auditingD. expecting

3.At last we got the letter we had long () expecting.A. areB. wereC. beD. been

4.Project work packages are typically decomposed into smaller components called activities to provide a basis for (72) ,scheduling, executing, and monitoring and controlling the project work.A．reviewingB．estimatingC．auditingD．expecting

更多“I’m expecting a letter____Jimmy. ”相关问题

  • 第1题：

    Project work packages are typically
    decomposed into smaller components called activities to provide a basis for （ ） , scheduling, executing, and
    monitoring and controlling the project work.

    A.reviewing
    B.estimating
    C.auditing
    D.expecting

    答案：B
    解析：

  • 第2题：

    22、假设整型变量n,m,i已经声明和赋值，且大于n＞2.有如下运算规则:如果n是素数，m=0；否则，m=n。如下程序中正确实现此运算的是？

    A．m=n; for(i=2;i＜n;i++) if(n%i==0){ m=0;break; }

    B．m=0; for(i=2;i＜n;i++) if(n%i==0){ m=n;break; }

    C．m=0; for(i=2;i＜n;i++) if(n%i==0){ m=n;continue; }

    D．m=n; for(i=2;i＜n;i++) if(n%i==0){ m=0;continue; }


    m=n; for(i=2;i if(n%i==0){ m=0;break; }

  • 第3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);


    D

  • 第4题：

    写出以下程序的运行结果： int main（) { int m,s,i; for（m=2;m＜10;m++) { s=0; for（i=1;i＜m;i++) if（（m%i)==0) s=s+i; if（s==m) printf（“%d”,m); } return 0; }


    10

  • 第5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);


    A

  • 第6题：

    假设整型变量n,m,i已经声明和赋值，且大于n＞2.有如下运算规则:如果n是素数，m=0；否则，m=n。如下程序中正确实现此运算的是？

    A．m=n; for(i=2;i＜n;i++) if(n%i==0){ m=0;break; }

    B．m=0; for(i=2;i＜n;i++) if(n%i==0){ m=n;break; }

    C．m=0; for(i=2;i＜n;i++) if(n%i==0){ m=n;continue; }

    D．m=n; for(i=2;i＜n;i++) if(n%i==0){ m=0;continue; }


    B解析：本程序中for循环的次数为c-1+1共循环了c次，而每循环一次s的值加1，所以s的值为s+c

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