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Start: Place dividend in remainder 第三种除法算法 Remainder Divisor 1. Shift the remainder register left 1 bit 000001110010 2. Subtract the Divisor register from the left half of the remainder register, place the result in the left half of the remainder register Remainder >=0 Test Remainder <0 maine 3a. Shift the 3b. Restore the original value by adding the divisor Remainder register register to the left half of the Remainderregister, to the left setting &place the sum in the left half of the remainder the new rightmost register Also shift the Remainder register to the bit to 1 left. setting the new least significant bit to o nth No: n repetitions etition Yes: n repetitions(n=4 here Done. shift left half of remainder right i bit. 北京大学计算机科学技术系 计算机系统结构教研室ñ¯M§¯æ*§cù ¯æù;étÐ@ \9ýÇ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