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第二种除法算法 Start: Place dividend in remainder Remainder Quotient Divisor 1. Shift the remainder register left i bit 0000011100000010 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 <o Remained 3a. Shift the 3b. Restore the original value by adding the divisor Quotient registerregister to the left half of the Remainder register, to the left setting &place the sum in the left half of the Remainder the new rightmost register. Also shift the Quotient register to the left, bit to 1 setting the new least significant bit to O th HNo:<repetitions epetition Yes: n repetitions(n=4 here) 北京大学计算机科学技术系 Done 计算机系统结构教研室ñ¯M§¯æ*§cù ¯æù;étÐ@ E5HVWRUHWKHRULJLQDOYDOXHE\DGGLQJWKH'LYLVRU UHJLVWHUWRWKHOHIWKDOIRIWKH5HPDLQGHUUHJLVWHU SODFHWKHVXPLQWKHOHIWKDOIRIWKH5HPDLQGHU UHJLVWHU$OVRVKLIWWKH4XRWLHQWUHJLVWHUWRWKHOHIW VHWWLQJWKHQHZOHDVWVLJQLILFDQWELWWR \¼ýÇ 5HPDLQGHU 4XRWLHQW 'LYLVRU 7HVW 5HPDLQGHU 5HPDLQGH 5HPDLQGHU U≥ 6XEWUDFWWKH'LYLVRUUHJLVWHUIURPWKH OHIWKDOIRIWKH5HPDLQGHUUHJLVWHU SODFHWKH UHVXOWLQWKHOHIWKDOIRIWKH5HPDLQGHUUHJLVWHU D6KLIWWKH 4XRWLHQWUHJLVWHU WRWKHOHIWVHWWLQJ WKHQHZULJKWPRVW ELWWR 6KLIWWKH5HPDLQGHUUHJLVWHUOHIWELW 'RQH <HV QUHSHWLWLRQV Q KHUH QWK UHSHWLWLRQ" 1R QUHSHWLWLRQV 6WDUW 3ODFH'LYLGHQGLQ5HPDLQGHU