for(i=1，t=1，s=0；i<=n；i++) {t=t*i；s=s+t；}的时间复杂度为O(n)。
A is helpful in looking for a job, but it doesn’t mean a job. （1.0分）
已知文本框Text1中输入了一篇英文短文，并编写了如下程序段：Str_x = Text1.Textn = Len(Str_x)m = 0t = 0For i = 1 To nw = UCase(Mid(Str_x, i, 1))If w >= A And w <= Z ThenIf t = 0 Then m = m + 1t = t + 1Elset = 0End IfNextPrint m该程序段的功能为统计并输出英文短文中________。
W: What can I do for you, sir?M: I’m looking for sports shoes.Size seven.Question: What does the man want to buy?
程序填空完成功能：求分数序列 2/1,3/2,5/3,8/5,13/8 …… 的前 20 项之和。 #include using namespace std; int main() { double i,n=1,m=1,t,s=0 ; for (i=1;i<=20;i++) { t = n ; n = m ; 【 】 ; s = s + m/n ; } cout<
以下程序的输出结果是#includeint main(void){int b[3][3]={0,1,2,0,1,2,0,1,2},i,j,t=1;for(i=0;i<3;i++)for(j=i;j<=i;j++)t=t+b[i][b[j][j]];printf(\%d\\n\,t);return 0;}
有以下程序inculde ＜stdio.h＞Main(){ int B[3][3]={0,1,2, 0,1,2,0,1,2},i,j,t+1;For (i=0;i＜3;i++)For (j=1;j＜=1;j++) t + =B[i]B[[j][i]];printF(“%d\n”,t);}程序运行后的输出结果是
听力原文：M: Look, I’m sorry I didn’t turn up for the match yesterday, but it wasn’t really my fault, you know.
下列程序段的时间复杂度为 。 for（i=0； i ＜m； i++) for（j="0；" j++) c[i][j]="c[i][j]+a[i][k]*b[k][j]；&lt;br/" for（i="0；" for（k="0；" k++)＞ A、O（m*n*t)
听力原文：M Is there anything that I can help you with？ Are you looking for something in particular or are you just browsing？
下列程序段的执行结果为______。Dim M（10), N（10)I=3For T = 1 To 5M（T) = TN（I) = 2 * I + TNext
阅读下列程序段：m = 0For i = 1 To 3For j = 1 To im = m + jNext j, i执行以上的循环后，i和m的值分别为()。
I have got a loaf of bread; now I' m looking for a knife______.
听力原文：W: I am looking for quality paper to type my essay. I don't see any on the shelf.M: I saw some in the stockroom this morning. I will go and check.Q: What does the woman want to buy?(5)A．A bookshelf.B．A typewriter.C．Some stocks.D．High-quality paper.
By this time next year,（), and I will already be looking for a job.
听力原文：M: Excuse me. I'm looking for this book. It's in the list of titles but I couldn't find it on the shelf.
It is he that I am looking for.（）
【单选题】下列程序段的时间复杂度为（）。 for（i=0；i＜m； i++) for（j=0； j＜t； j++) c[i][j]=0； for（i=0；i＜m； i++) for（j=0；j＜t； j++) for（k=0；k＜n； k++) c[i][j]=c[i][j]+a[i][k]*b[k][j]；
分析程序的上界O和下界W。 for i = 0 to m M[0, i] = id for j = 0 to n M[j, 0] = jd for i = 1 to m for j = 1 to n M[i, j] = min(a[xi, yj] + M[i-1, j-1], d + M[i-1, j], d + M[i, j-1]) return M[m, n] 该程序时间复杂度的上界是O（____）、下界是W（_____）。
s=0 for i in range（1,5): m=1 for j in range（1,i+1): m=m*j s=s+m print（s) 程序运行结果是：（）
I’m good ______ old peoplend I’m good ______ plying the guitr, so I cn join you to help them.I’m good ______ old peoplend I’m good ______ plying the guitr, so I cn join you to help them.for, with B．for,t C．with,t D．t, for
—I’m still looking for a topic for my assignment. And I don’t want to do too much research. —___ _ choosing a current issue There is a lot happening in the region at present.
Part 1 I’m Looking for the Museum. Listen to the conversation and circle the letter of the correct answer. Where are the speakers now？
完成下方代码填空。斐波纳契数列有如下特点：第1、2个数都是1，从第三个数开始，每个数都是前两个数之和。下列程序的功能是求数列的前m（m＞1）个数，按每行5个数输出： class Program { static void Main（string[] args) { int f1 = 1, f2 = 1, m; m = Convert.ToInt16（Console.ReadLine（)); Console.Write（f1 + "t" + f2 + "t"); for （int i = 3; i ＜= m; i++) { f2 = f1 + f2; f1 = ___1___; Console.Write（___2___ + "t"); if （i % ___3___ == 0) Console.WriteLine（""); } } }
