此题为判断题(对,错)。
第1题:
A.should have helped
B.would help
C.would have helped
D.helped
第2题:
A. would have seen
B. would see
C. have seen
D. saw
第3题:
30、如下程序的输出结果是 int main() { char books[][20]={"English","Math","Physical"}; int i,j; for(i=0;i<3;i++) { for(j=0;books[i][j]!=0;j++){ if(books[i][0]<books[i][j]) books[i][0]= books[i][j]; } } printf("%c",books[0][0]); return 0; }
A.E
B.s
C.n
D.h
第4题:
I’m looking forward _______ you in the near future.
A、to seeing
B、to see
C、seeing
D、seen
第5题:
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<j,k>表示,则Ee[I] = Ve[j];
d. 边活动最晚开始时间 El[I], 若边I由<j,k>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;
第6题:
29、如下程序的输出结果是 int main() { char books[][20]={"English","Math","Physical"}; int i,j; for(i=0;i<3;i++) { strcat(books[i],"book"); } printf("%s",&books[i-1][3]); return 0; }
A.Physicalbook
B.sical
C.Physical
D.sicalbook