I've got the order from the boss that the work______finished before 5p. m. today.
A．would be
B．be
C．will have been
D．could be
I’ll give him your message___he comes back from lunch.A. untilB. as soon asC. whileD. before
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根据以下材料回答第 1～5题：A．I'm leaving．B．Thank youC．I'll do my bestD．and sit hereE．Please sit downF．But l worry aboutG．Take it easyH．Don’t forget第 56 题 Boss：Come in，please．Oh，Mary，come over 56 ．Your annual report is well done．Mary： 57Boss：I know you’re a capable person．Mary：Thank you for saying that． 58 the wrong figures I gave you last time．Boss： 59 ．Everyone makes mistakes．Mary：Thank you SO much for your forgiveness． 60．
●试题四下列给定程序中，函数fun()的功能是：对N名学生的学习成绩，按从高到低的顺序找出前m(m≤10)名学生来，并将这些学生数据存放在一个动态分配的连续存储区中，此存储区的首地址作为函数值返回。注意：部分源程序给出如下。请勿改动主函数main和其他函数中的任何内容，仅在函数fun()的横线上填入所编写的若干表达式或语句。试题程序：#include#include#include#include#define N 10typedef struct ss{char num［10］;int order;}STU;STU *fun(STU a［］, int m){STU b［N］,*tt;int i,j,k;(1) ;for(i=0; i10){printf("＼nGive the number of thestudets who have better score:");scanf("%d",&m);}p_order=fun(a,m);printf("*** THE RESULT ***＼n");printf("*** The top students ***＼n");for(i=0; i<m; i++)printf(" %s %d＼n",p_order［i］.num,p_order［i］.order);free(p_order);}
Boss: Come in, please. Oh, Mary, come over (56) Your annual report is well done.Mary: (57)Boss: I know you're a capable person.Mary: Thank you for saying that. (58) the wrong figures I gave you last time.Boss: (59) Everyone makes mistakes.Mary: Thank you so much for your forgiveness. (60)56.A. I'm leavingB. Thank youC. I'll do my bestD. and sit hereE. Please sit downF. But I worry aboutG. Take it easyH. Don't forget
对由SELECT—FROM—WHERE—GROUP—ORDER组成的SQL语句，其在被DBMS处理时，各子句的执序次序(43)。A．SELECT—FROM—GROUP—WHERE—ORDERB．FROM—SELECT—WHERE—GROUP—ORDERC．FROM—WHERE—GROUP—SELECT—ORDERD．SELECT—FROM—WHERE—GROUP—ORDER
在SELECT语句中，用来指定查询所用的表的子句是A）WHERE B）GROUP BY C）ORDER.BY D）FROM
B 宽度优先（种子染色法）5．关键路径几个定义： 顶点1为源点，n为汇点。a. 顶点事件最早发生时间Ve[j], Ve [j] = max{ Ve [j] + w[I,j] },其中Ve (1) = 0;b. 顶点事件最晚发生时间 Vl[j], Vl [j] = min{ Vl[j] – w[I,j] },其中 Vl(n) = Ve(n);c. 边活动最早开始时间 Ee[I], 若边I由表示，则Ee[I] = Ve[j];d. 边活动最晚开始时间 El[I], 若边I由表示，则El[I] = Vl[k] – w[j,k];若 Ee[j] = El[j] ，则活动j为关键活动，由关键活动组成的路径为关键路径。求解方法：a. 从源点起topsort,判断是否有回路并计算Ve;
