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第一种除法算法 Start: place dividend in remainder 对n位商和余数,需要n+1步. 1. Subtract the Divisor register from the 余数 商 除数 Remainder register. and place the result 0000 0111 0000 0010 0000 Lin the Remainder register. Remainder≥0 Test Remainder <o remainder 2a. shift the 2b Restore the original value by adding the Quotient register Divisor register to the Remainder register, to the left setting place the sum in the Remainder register. Also the new rightmost shift the Quotient register to the left, setting bit to 1 the new least significant bit to 0 3. Shift the divisor register right 1 bit n+1 No:<n+repetitions epetition Yes: n+l 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@ 6KLIWWKH'LYLVRUUHJLVWHUULJKWELW E 5HVWRUHWKHRULJLQDOYDOXH E\DGGLQJWKH 'LYLVRUUHJLVWHUWRWKH5HPDLQGHUUHJLVWHU SODFHWKHVXPLQWKH5HPDLQGHUUHJLVWHU$OVR VKLIWWKH4XRWLHQWUHJLVWHUWRWKHOHIWVHWWLQJ WKHQHZOHDVWVLJQLILFDQWELWWR 'RQH \0ýÇ 6WDUW 3ODFH'LYLGHQGLQ5HPDLQGHU eÍ Q! `-DÈÔUQ9 -D 8D 7HVW 5HPDLQGHU 5HPDLQGHU≥ 5HPDLQGHU 6XEWUDFW WKH'LYLVRUUHJLVWHUIURPWKH 5HPDLQGHUUHJLVWHUDQGSODFHWKHUHVXOW LQWKH5HPDLQGHUUHJLVWHU D 6KLIWWKH 4XRWLHQWUHJLVWHU WRWKHOHIWVHWWLQJ WKHQHZULJKWPRVW ELWWR <HV QUHSHWLWLRQV Q KHUH Q UHSHWLWLRQ" 1R QUHSHWLWLRQV