若s是int型变量,且s=6,则表达式s&&2+(s+1)%2的值是____________。
若s是int型变量,且s=6,则表达式s&&2+(s+1)%2的值是____________。
2.15类双绞线(CAT5)的最高传输速率为() A: 250Mbit/s B: 100Mbit/s  C: 155Mbit/s D: 600Mbit/s 
2.15类双绞线(CAT5)的最高传输速率为() A: 250Mbit/s B: 100Mbit/s  C: 155Mbit/s D: 600Mbit/s 
若有定义int x;char s[20];正确的输入语句是( )。 A: scanf("%d %c",&x,&s); B: scanf("%d %c",&x,s); C: scanf("%d %s",&x,&s); D: scanf("%d %s",&x,s);
若有定义int x;char s[20];正确的输入语句是( )。 A: scanf("%d %c",&x,&s); B: scanf("%d %c",&x,s); C: scanf("%d %s",&x,&s); D: scanf("%d %s",&x,s);
有定义语句:intb;charc[10];,则正确的输入语句是()。 A: scanf("%d%s",&b,&c); B: scanf("%d%s",&b,c); C: scanf("%d%s",b,c); D: scanf("%d%s",b,&c);
有定义语句:intb;charc[10];,则正确的输入语句是()。 A: scanf("%d%s",&b,&c); B: scanf("%d%s",&b,c); C: scanf("%d%s",b,c); D: scanf("%d%s",b,&c);
对于集合S和T,下列不属于集合类型的操作是() A: S-T B: S+T C: S&T D: S^T
对于集合S和T,下列不属于集合类型的操作是() A: S-T B: S+T C: S&T D: S^T
下面程序的功能是从键盘输入一串字符,统计其中有多少个单词,单词之间用空格分隔。则【1】中应选择的是。 A: qian==’˽’&&s[i]!=’˽’ B: qian==’˽’&&s[i]==’˽’ C: qian!=’˽’&&s[i+1]!=’˽’. D: qian!=’˽’&&s[i]!=’˽’
下面程序的功能是从键盘输入一串字符,统计其中有多少个单词,单词之间用空格分隔。则【1】中应选择的是。 A: qian==’˽’&&s[i]!=’˽’ B: qian==’˽’&&s[i]==’˽’ C: qian!=’˽’&&s[i+1]!=’˽’. D: qian!=’˽’&&s[i]!=’˽’
执行以下代码,运行结果defsplit(s):returns.split("t")s="Happybirthdaytoyou!"print(split(s)) A: 运行出错 B: ["&quot","Happy&quot",null,"&quot","birthday&quot",null,"&quot","to&quot",null,"&quot","you!&quot"] C: ["Happybir","&nbsp","hday","&nbsp","oyou!"] D: ["Happy","&nbsp","bir","&nbsp","hday","&nbsp","o","&nbsp","you!"]
执行以下代码,运行结果defsplit(s):returns.split("t")s="Happybirthdaytoyou!"print(split(s)) A: 运行出错 B: ["&quot","Happy&quot",null,"&quot","birthday&quot",null,"&quot","to&quot",null,"&quot","you!&quot"] C: ["Happybir","&nbsp","hday","&nbsp","oyou!"] D: ["Happy","&nbsp","bir","&nbsp","hday","&nbsp","o","&nbsp","you!"]
Allen Winkler, CFA, recently had lunch with Kim Thompson, a former professor of his, who told him of a new valuation model she had developed. Winkler recreated Thompson's model with some revisions and back-tested it using data provided by Standard & Poor's (S&P) with impressive results. Winkler's firm launches a mutual fund based on the revised model, and Winkler provides a discussion of the principles underlying the model and the test results. Is Winkler required to credit Thompson for having developed the model and S&P as the source of the data? A: Both of these sources must be cited. B: Neither of these sources must be cited. C: Only one of these sources must be cited.
Allen Winkler, CFA, recently had lunch with Kim Thompson, a former professor of his, who told him of a new valuation model she had developed. Winkler recreated Thompson's model with some revisions and back-tested it using data provided by Standard & Poor's (S&P) with impressive results. Winkler's firm launches a mutual fund based on the revised model, and Winkler provides a discussion of the principles underlying the model and the test results. Is Winkler required to credit Thompson for having developed the model and S&P as the source of the data? A: Both of these sources must be cited. B: Neither of these sources must be cited. C: Only one of these sources must be cited.
有声明语句:int a,b;char c[10]; ,则正确的输入语句是 A: scanf("%d%d%s",&a,&b,c); B: scanf("%d%d%s",&a,&b,&c); C: scanf("%d%d%s",a,b,c); D: scanf("%d%d%c",&a,&b,&c);
有声明语句:int a,b;char c[10]; ,则正确的输入语句是 A: scanf("%d%d%s",&a,&b,c); B: scanf("%d%d%s",&a,&b,&c); C: scanf("%d%d%s",a,b,c); D: scanf("%d%d%c",&a,&b,&c);
\(A\)同上题,将其对角化\(A=S\Lambda S^{-1}\)的方阵\(S\)可以是 A: \(\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}\) B: \(\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\) C: \(\begin{pmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix}\)
\(A\)同上题,将其对角化\(A=S\Lambda S^{-1}\)的方阵\(S\)可以是 A: \(\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}\) B: \(\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\) C: \(\begin{pmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix}\)