1班 100240053 吴佳桢 已知： L0=R0=“01100001011000100110001101100100” K1=“011000010110001001100011011001000110010101100110”。 求：R1? R1=L0⊙f(R0,K1) 0 0 1 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 1 1 0 1 0 1 1 0 0 0 0 1 0 0 0 R0异或K1=010100010100100101100111010101000000111001101110 令：列=D2D3D4D5 行=D1D6 S1=6 S2=2 S3=13 S4=6 S5=15 S6=12 S7=14 S8=2 →0110 0010 1101 0110 1111 1100 1110 0010 结果为：10100000000101001001111110000110 所以R1=10100000000101001001111110000110 1班 100240053 吴佳桢 本文来源：https://www.wddqw.com/doc/1b3cda322179168884868762caaedd3382c4b540.html