• 2022-06-15
    求极限:[tex=5.929x2.5]ENxIatiC2yqgaopSQCG83v55B1G0LTFPU12wzUQ73Dn7pPRJ32Lq74NvGQEumohxBboKRAKFri41daoY6QvOZw==[/tex]
  • 解:[tex=23.571x4.5]9N1RyOEuBqasOARUdep0scS4NOgnst/A6WBiaecYxWX4IMytRhRc7hwPCOHKSbLE2uz1M1Y8/Gxd0mQfEE5iV1p1cU3Px3uB+MEeiLkEee30ouLfjCHHwC+VkCtRPFWclmWE0eC6NRFOuCgJr5nxuPrKQQA+AHCcGUd4N87BZ10jAlve6r/9C83rcnnAeJ0kwRCxHaoRjFf57flMmcNY9WjByWuq90JxTG+DvE/2WOxGBubWlMDNPZ828M1rF66UEBslKYH02yNdKtiUzvoirz+ehZycA3zymh1XQddsIyHxR7pfkBEqh0W2NU6ybjzAB2uboBrRJi2YnHaRKCcPP5NRuy3XEPSdnGk/8r0CutUVuzlPb8GyVoDP56cd/bda[/tex]

    内容

    • 0

      求下列函数的导函数:(1) [tex=5.0x2.357]X/CieCDGJ7iPQ3YFWuscHxHrcIE/dPFa9tFyiJXze8A=[/tex](2)[tex=6.643x1.714]Oj74y/L+OxY81QME5JWMcl+7PZ2FGQswwvjgVhjq1Dmb6dBU0oAjZBW7eFBVjqo6[/tex]

    • 1

      void Sort ( int iRecordNum, int iType ) 1 { 2 int x=0; 3 int y=0; 4 while ( iRecordNum > 0 ) 5 { 6 if ( iType==0 ) 7 x=y+2; 8 else 9 if ( iType==1 ) 10 x=y+10; 11 else 12 x=y+20; 13 } 14 } 要求(1)给以上代码画出控制流图(2)控制流图的环复杂度V(G),写出独立路径

    • 2

      下列程序的输出结果是 。using namespce std; #include &#91;iostream&#93; int main() { int x=1,i=1; for (; x &#91; 50; i++) { if(x &#93;= 10) break; if(x % 2 != 0){ x += 3; continue; } x-=-1; } cout<<x<<' '<<i<<endl; return 0; } A: 12 7 B: 11 6 C: 12 6 D: 11 7

    • 3

      求下列极限:(1)lim(x→0)(∫x0ln(1+2t^2)dt/x^3

    • 4

      采用基2时间抽取FFT算法流图计算8点序列的DFT,第一级的数据顺序为 A: x[0],x[2],x[4],x[6],x[1],x[3],x[5],x[7] B: x[0],x[1],x[2],x[3],x[4],x[5],x[6],x[7] C: x[0],x[4],x[2],x[6],x[1],x[5],x[3],x[7] D: x[0],x[2],x[1],x[3],x[4],x[6],x[5],x[7]