• 2022-05-28
    设 [tex=9.786x1.286]DQ1ZsnXfTUHqv3aeNI8TmtYnGbUL/0+sIYsjPWXY8gvgx6R0cuYt0RtG7mXTbZvLu6NAcQR1sSciIWMyvqLruQ==[/tex]. 令 [tex=2.0x1.214]L13v13gszAHTYXVyhXnbMg==[/tex] 分别为[tex=19.429x7.643]apkvZnpOyWUQAwLC3506svu/wi6+Vf6yufHs2t0U/LSrh/UR0yRmjSDCWbIkiRP5KNj0dZklPPpanqGHBxZxFUke2yxSRM0Le7aQ3x9IDgx2oqCjLeQmJ9I9Lm9FZQ1WrcN8ZLFliV7YbEntpFqEeNpUcj5kCwUcku7rrZhd6wBL1Aql4yV6U060EGb17OYbw4qmpXPgwT1S+bq2Xxvj+gLg79Xtu9OzAY2CGBuSFEPGuNht6ymFtD2ttg3bfXZG/8v7eXiI/MgwKhieqFXTVaxYoXk+3j4luucBI8n4MULK8T67TezHoDNdox2qVCzjtZTPjtIgl2EZZEy9x/2N7w==[/tex],[tex=23.786x7.714]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[/tex].1) 计算 [tex=1.571x1.0]KtUvskt9CnZGjwvAZvfMbg==[/tex].2) 求 [tex=2.357x1.0]NovbxKl63Ey/milqTcbe/0+weZrZwV0IV2WDP9e/d4g=[/tex].3) [tex=9.143x1.214]W+JKlZTCQuMKHwApS0vbt6P88J6iGPqmse/zfvqGaOiRTh3wV751xrP0M0uf4DMSNyBUI4gfEk1UxitcE2tRlw==[/tex] 时,[tex=3.143x1.0]NovbxKl63Ey/milqTcbe/8jdjiYrbSNvYzAkW4iNzpM=[/tex]?
  • 解 1) [tex=11.0x4.643]EoTVfHzIFrQhWLWnT90KgNkNQca5j62W8+u+NZHcjPS6bPrWPH29O4xMuPsHYl3kJVE/4RLP0pbgcylPEEaTg55EuQoN/5NDFouKEw7BiAkdOjkKgmPQqOe9b4sO75IoCd1qg7BnWKfHmNzTTGz1eA8au4O1JM2EOxeLHOUm2aXktMvST3ogvsT51vAhlXSZNWFD96UwoWG9eCdacd1AZA==[/tex],其中 [tex=10.357x1.214]n9xhw4nPuMOemBTyzHZOHKWiQLDFg9XkcJeg09z+3ldVTOqRhcpbofe7i9eqxbO/UOJfLOf3c9H8J5StdDrB7g==[/tex].2)首先注意 [tex=14.786x1.286]0sI4VLUEKvUp2doVE1sIHJpBFXLUWCgHOwsEPrzhM3sMESqpdAaKRcEPOZ3Cygy3y9axdeON6Y2K2B304i+aYmPBoMYmvyXzpr70R8fEArQ=[/tex],[tex=15.0x1.357]V1D753We7vezsBlKQyfrUtPHRuPVJHnTFx7zxy2GVEUkCAUEDERCZ79qG+8gEeWa+EySrgH+z+QkpwlPABFnCs+qy2JTVjRNwp+UFgWQZ+0DSig3l3JzMsmVK+VzC67FDH1A9CyQRJ0M3VDWzBo7XA==[/tex] 因此[p=align:center][tex=18.929x2.429]V1D753We7vezsBlKQyfrUjx/wz90fnxOVb9iRbCPFiVwmcntTS/u6Mgybiz7Vuhp/J+/xC71SowzVwh/iaJ2T+zJcpGP8Omgl7Xdla/uCQUl498H0/kFyd/E2WhFmQ/j8df0UsB3BeuANoMwE8RrYbOfUX0pqPEdtfQptU/w5l6d1vMhLNVeTOPcC+lrZ5ub4qPYLuqVeuj77mL9OyaMPA==[/tex].3)将 [tex=2.0x1.214]vnzjVhyzo/NIhVUgFyjLlA==[/tex] 中的 [tex=0.786x1.0]k2s61G+KZI8Yw0ah0O7QEQ==[/tex] 代之以 [tex=2.143x1.214]8MBdgAWt5nqC9U1ZrkjptQ==[/tex],所得矩阵记为 [tex=5.0x1.357]pb1eN5lKXjopcyO75VKCypnVZgJtV+SaOPduG0FQLjE=[/tex] 再令[p=align:center][tex=18.714x1.357]ZvkGQ14d5y3azajfNLU8uDGzGbCwnhZoAYfcsL5687eNzwI7SqSBKGCJd7mrHXvLM1pAmnrPhkRHTHPwlMew6LQVS7fcDwHaSvocPsjbG6BBxDfyPwLcm2ULzovf+DVLhQuYBkj31IosZWzkPt+jmqPv5Hze4JrXJhLUViI/CAk=[/tex].这是 [tex=0.643x1.0]+D9NhKovEP8INGz+KZnr1A==[/tex] 的 [tex=0.643x0.786]/he/ol8BkDuTTL9yMPtH4Q==[/tex] 次多项式,至多 [tex=0.643x0.786]/he/ol8BkDuTTL9yMPtH4Q==[/tex] 个零点,0 是一个零点. 注意 [tex=3.714x1.357]NovbxKl63Ey/milqTcbe/xBsy61JrD2QZ4gY5pRW/OuVoJZrWaCpveo3LeVzi/rg[/tex] 也是 [tex=0.643x1.0]+D9NhKovEP8INGz+KZnr1A==[/tex] 的多项式,且当 [tex=3.714x1.286]Cfove8/8JZ+FT9G/qWRm4tGMspifW48h8ex6ujlxrb8=[/tex] 时,根据本题的 1) 与 2),有[p=align:center][tex=7.714x1.5]V1D753We7vezsBlKQyfrUpObS25+e5Oc5U5yTyc93eDdwunTY8RzuoCkL7CUjuTM[/tex].等式两边的多项式在无穷多点相等,故二多项式相等,于是[p=align:center][tex=12.071x1.5]V1D753We7vezsBlKQyfrUqv4Oo6cPfhaSNZkNZFP4mcU6tr4ivsnT7BlQ5BpQpGNtrcb1n/NlOQYj4A8EsYqtQ==[/tex].

    内容

    • 0

      下面是图的拓扑排序的是?(多选)[img=340x240]1802faef6ebcc2a.png[/img] A: 2 8 0 7 1 3 5 6 4 9 10 11 12 B: 2 8 7 0 6 9 11 12 10 1 3 5 4 C: 8 2 7 3 0 6 1 5 4 9 10 11 12 D: 8 2 7 0 6 9 10 11 12 1 3 5 4

    • 1

      【单选题】myarray1=np.arange(15) myarray2=myarray1.reshape(5,3) print( myarray1) print(myarray2) 输出值是? A. [ 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15] [[ 0  1  2] [ 3  4  5] [ 6  7  8] [ 9 10 11] [12 13 14]] B. [ 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15] [[ 1  2  3] [ 4  5  6] [ 7  8  9] [10 11 12] [13 14 15]] C. [ 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14] [[ 0  1  2] [ 3  4  5] [ 6  7  8] [ 9 10 11] [12 13 14]] D. [ 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14] [[ 1  2  3] [ 4  5  6] [ 7  8  9] [10 11 12] [13 14 15]]

    • 2

      说明S盒变换的原理,并计算当输入为110101时的S1盒输出。 [br][/br] n\m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 S1 0 14 4 13 1 2 15 11 8 3 10 6 12 5 9 0 7 1 0 15 7 4 14 2 13 1 10 6 12 11 9 5 3 8 2 4 1 14 8 13 6 2 11 15 12 9 7 3 10 5 0 3 15 12 8 2 4 9 1 7 5 11 3 14 10 0 6 13

    • 3

      list(range(-1,10,2)) A: [0, 2, 4, 6, 8, 10] B: [-1, 1, 3, 5, 7, 9] C: [0, 2, 4, 6, 8]

    • 4

      设DES加密算法中的一个S盒为: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 1 2 3 14 4 13 1 2 15 11 8 3 10 6 12 5 9 0 7 0 15 7 4 14 2 13 1 10 6 12 11 9 5 3 8 4 1 14 8 13 6 2 11 15 12 9 7 3 10 5 0 15 12 8 2 4 9 1 7 5 11 3 14 10 0 6 13 若给定输入为101101,则该S盒的输出的二进制表示为