• 2022-05-31 问题

    39号元素钇的核外电子排布式是下列排布中的( ) A: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 4 d 1 5 s 2 B: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 5 s 2 5 p 1 C: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 4 d 2 5 s 1 D: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 5 s 2 5 p 1

    39号元素钇的核外电子排布式是下列排布中的( ) A: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 4 d 1 5 s 2 B: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 5 s 2 5 p 1 C: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 4 d 2 5 s 1 D: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 5 s 2 5 p 1

  • 2022-06-10 问题

    电子组态1s2p所构成的原子态为( )。 A: 1s2p1S0,1s2p1P1, 1s2p3S1 1s2p3P2,1,0 B: 1s2p1S0,1s2p1P1 C: 1s2p1S0,1s2p3S1 D: 1s2p1P1,1s2p3P2,1,0

    电子组态1s2p所构成的原子态为( )。 A: 1s2p1S0,1s2p1P1, 1s2p3S1 1s2p3P2,1,0 B: 1s2p1S0,1s2p1P1 C: 1s2p1S0,1s2p3S1 D: 1s2p1P1,1s2p3P2,1,0

  • 2022-05-29 问题

    试判断原子态:1s1s3S1,1s2p3P2,1s2p1D1,2s2p3P2中,下面哪组是存在的? A: 1s1s3S1,1s2p3P2; B: 1s2p3P2,1s2p1D1; C: 1s2p3P2,2s2p3P2; D: 1s1s3S1,1s2p1D1;

    试判断原子态:1s1s3S1,1s2p3P2,1s2p1D1,2s2p3P2中,下面哪组是存在的? A: 1s1s3S1,1s2p3P2; B: 1s2p3P2,1s2p1D1; C: 1s2p3P2,2s2p3P2; D: 1s1s3S1,1s2p1D1;

  • 2022-06-06 问题

    高P/S值是指() A: P/S≥1 B: P/S≤1 C: P/S>1 D: P/S>0.5

    高P/S值是指() A: P/S≥1 B: P/S≤1 C: P/S>1 D: P/S>0.5

  • 2022-06-14 问题

    假如集合 S={ 1, 2, 3, 4 },后面 5 个命题中,哪些是正确的?ⅰ) { {2} }⊆P(S),ⅱ) { 1, 3 }⊆P(S),ⅲ) { 1, 3 }∈P(S),ⅳ) { {2, 4} }⊆P(S),ⅴ) { 4 }⊆P(S)

    假如集合 S={ 1, 2, 3, 4 },后面 5 个命题中,哪些是正确的?ⅰ) { {2} }⊆P(S),ⅱ) { 1, 3 }⊆P(S),ⅲ) { 1, 3 }∈P(S),ⅳ) { {2, 4} }⊆P(S),ⅴ) { 4 }⊆P(S)

  • 2022-06-07 问题

    对于文法G(S'),该文法识别活前缀的DFA如下图,状态I2包含的项目有G(S'):(0) S' → S(1) S → Pa(2) S → Pb(3) S → c(4) P → Pd(5) P → Se(6) P → f[img=3601x1590]1803958e6ecda94.png[/img] A: S → PŸa B: S → PŸb C: S → PŸc D: S → PŸd

    对于文法G(S'),该文法识别活前缀的DFA如下图,状态I2包含的项目有G(S'):(0) S' → S(1) S → Pa(2) S → Pb(3) S → c(4) P → Pd(5) P → Se(6) P → f[img=3601x1590]1803958e6ecda94.png[/img] A: S → PŸa B: S → PŸb C: S → PŸc D: S → PŸd

  • 2022-06-27 问题

    下列程序段没有错误的是( )。 A: int*p;cin>>*p; B: int*s,k;*s=100 C: int *s ,k; char *p ,c; s=&k; p=&c; *p='a'; *s=1; D: int *s,k; char *p,c; s=&k; p=&c; s=p;

    下列程序段没有错误的是( )。 A: int*p;cin>>*p; B: int*s,k;*s=100 C: int *s ,k; char *p ,c; s=&k; p=&c; *p='a'; *s=1; D: int *s,k; char *p,c; s=&k; p=&c; s=p;

  • 2022-07-29 问题

    在循环双链表的p结点之后插入s结点的操作是______。 A: p→next=s;p→next→prior=s;S→prior=p;S→next=p→next; B: s→next=p;s→next=p→next;p→next=s;p→next→prior=s; C: p→next=s;s→prior=p;p→next→prior=s;s→next=p→next; D: s→prior=p;s→next=p→next;p→next→prior=s;p→next=S;

    在循环双链表的p结点之后插入s结点的操作是______。 A: p→next=s;p→next→prior=s;S→prior=p;S→next=p→next; B: s→next=p;s→next=p→next;p→next=s;p→next→prior=s; C: p→next=s;s→prior=p;p→next→prior=s;s→next=p→next; D: s→prior=p;s→next=p→next;p→next→prior=s;p→next=S;

  • 2022-07-29 问题

    在循环双链表的p所指结点之后插入s所指结点的操作是() A: P—>next=s; B: p—>next=s;s—>prior=p;p—>next—>prior=s;p—>next—>prior=s;s—>prior=p;s—>next=p—>next;s—>next=p—>next C: s—>prior=p; D: s—>prior=p;s—>next=p—>next;s—>next=p—>next;p—>next=s;p—>next—>prior=s;p—>next—>prior=s;p—>next=s;

    在循环双链表的p所指结点之后插入s所指结点的操作是() A: P—>next=s; B: p—>next=s;s—>prior=p;p—>next—>prior=s;p—>next—>prior=s;s—>prior=p;s—>next=p—>next;s—>next=p—>next C: s—>prior=p; D: s—>prior=p;s—>next=p—>next;s—>next=p—>next;p—>next=s;p—>next—>prior=s;p—>next—>prior=s;p—>next=s;

  • 2021-04-14 问题

    完成在双循环链表结点p之后插入s的操作是()A.()p->next=s();()s()->()prior=p;()p()->()next()->()prior=s();()s()->()next=p()->()next;()B.()p()->()next()->()prior=s;()p()->()next=s;()s()->()prior=p;()s()->()next=p()->()next;()C.()s()->()prior=p;()s()->()next=p()->()next;()p()->()next=s;()p()->()next()->()prior=s();()D.()s()->()prior=p;()s()->()next=p()->()next;()p()->()next()->()prior=s();()p()->()next=s;

    完成在双循环链表结点p之后插入s的操作是()A.()p->next=s();()s()->()prior=p;()p()->()next()->()prior=s();()s()->()next=p()->()next;()B.()p()->()next()->()prior=s;()p()->()next=s;()s()->()prior=p;()s()->()next=p()->()next;()C.()s()->()prior=p;()s()->()next=p()->()next;()p()->()next=s;()p()->()next()->()prior=s();()D.()s()->()prior=p;()s()->()next=p()->()next;()p()->()next()->()prior=s();()p()->()next=s;

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