对于定域子系统, 某种分布所拥有的微观状态数[tex=1.571x1.214]5/6KlL+4s4+76HQrLLRoVA==[/tex]为[input=type:blank,size:4][/input]
未知类型:{'options': ['\xa0[tex=6.357x2.714]+s8mNtPUa0KuzYFUx1gmONm3wu1xKb1TjKyHP63I9w2xjYHCFs65yQfVWmk3J4RJj/pDOOo/RESwH/Fe1LM/KA==[/tex]', '\xa0[tex=6.0x2.571]+s8mNtPUa0KuzYFUx1gmOBrzZKIsaH3eDph4VFd2/t8XYQtFpWFVX+ohpZbN7yAQ[/tex][br][/br]', '\xa0[tex=5.786x2.786]lvRpyxr8CyqVudWLMIUbRkjVy8Kw2PKHuZ/Mr1WZvXp8Jw0KzwiVZU0dqhMcADE3[/tex][br][/br]', '[tex=4.714x2.786]Mitty9ZiBQjKZ+7oDxlYrWkm/1xs6yl2RiDfVsyUMAMkBZcm5N64lBS/yxUabA+4Ao7mavb4lDq2lN6TKu5Xlw==[/tex][br][/br]'], 'type': 102}
未知类型:{'options': ['\xa0[tex=6.357x2.714]+s8mNtPUa0KuzYFUx1gmONm3wu1xKb1TjKyHP63I9w2xjYHCFs65yQfVWmk3J4RJj/pDOOo/RESwH/Fe1LM/KA==[/tex]', '\xa0[tex=6.0x2.571]+s8mNtPUa0KuzYFUx1gmOBrzZKIsaH3eDph4VFd2/t8XYQtFpWFVX+ohpZbN7yAQ[/tex][br][/br]', '\xa0[tex=5.786x2.786]lvRpyxr8CyqVudWLMIUbRkjVy8Kw2PKHuZ/Mr1WZvXp8Jw0KzwiVZU0dqhMcADE3[/tex][br][/br]', '[tex=4.714x2.786]Mitty9ZiBQjKZ+7oDxlYrWkm/1xs6yl2RiDfVsyUMAMkBZcm5N64lBS/yxUabA+4Ao7mavb4lDq2lN6TKu5Xlw==[/tex][br][/br]'], 'type': 102}
举一反三
- set1 = {x for x in range(10)} print(set1) 以上代码的运行结果为? A: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} B: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10} C: {1, 2, 3, 4, 5, 6, 7, 8, 9} D: {1, 2, 3, 4, 5, 6, 7, 8, 9,10}
- >>>x= [10, 6, 0, 1, 7, 4, 3, 2, 8, 5, 9]>>>print(x.sort()) 语句运行结果正确的是( )。 A: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] B: [10, 6, 0, 1, 7, 4, 3, 2, 8, 5, 9] C: [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0] D: ['2', '4', '0', '6', '10', '7', '8', '3', '9', '1', '5']
- 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
- 已知S盒如下表,若输入为100010,则二进制输出为( ) [br][/br] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 7 13 14 3 0 6 9 10 1 2 8 5 11 12 4 15 1 13 8 11 5 6 15 0 3 4 7 2 12 1 10 14 9 2 10 6 9 0 12 11 7 13 15 1 3 14 5 2 8 4 3 3 15 0 6 10 1 13 8 9 4 5 11 12 7 2 14 A: 0110 B: 1001 C: 0100 D: 0101
- 【计算题】5 ×8= 6×4= 7×7= 9×5= 2×3= 9 ×2= 8×9= 7×8= 5×5= 4×3= 5+8= 6 ×6= 3×7= 4×8= 9×3= 1 ×2= 9×9= 6×8= 8×0= 4×7=