遺傳算法解決TSP問題 已知n個(gè)城市之間的相互距離，現(xiàn)有一個(gè)推銷員必須遍訪這n個(gè)城市，并且每個(gè)城市只能訪問一次，最后又必須返回出發(fā)城市。如何安排他對(duì)這些城市的訪問次序，可使其旅行路線的總長(zhǎng)度最短？
上傳時(shí)間： 2013-12-26
上傳用戶：trepb001
crc16校驗(yàn)子程序，采用查表法計(jì)算。 對(duì)于匯編開發(fā)人員比較有益。
上傳時(shí)間： 2013-12-03
上傳用戶：aig85
產(chǎn)生N=100個(gè)在（0,50）區(qū)間內(nèi)滿足對(duì)數(shù)正態(tài)分布的隨機(jī)數(shù)
標(biāo)簽： 100 50 對(duì)數(shù) 分布
上傳時(shí)間： 2014-01-01
上傳用戶：sjyy1001
產(chǎn)生N=100個(gè)在（0,50）區(qū)間內(nèi)滿足瑞利分布的隨機(jī)數(shù)
標(biāo)簽： 100 50 分布 隨機(jī)數(shù)
上傳時(shí)間： 2014-01-06
上傳用戶：han_zh
v n vb
標(biāo)簽： vb
上傳時(shí)間： 2015-10-01
上傳用戶：as275944189
生成Bipartite Graphs ./distributions -u -m 1 -M 10 -n 100 -s 500 > top_distrib ./distributions -p -2.2 -m 1 -M 100 -n 200 -s 500 > bottom_distrib ./random_bipartite -t top_distrib -b bottom_distrib > bn_test
標(biāo)簽： distributions top_distrib Bipartite Graphs
上傳時(shí)間： 2015-10-02
上傳用戶：yy541071797
多項(xiàng)式為x^8+x^5+x^4+1的CRC（循環(huán)冗余校驗(yàn)碼）的FPGA設(shè)計(jì)。
標(biāo)簽： FPGA CRC 多項(xiàng)式 循環(huán)冗余
上傳時(shí)間： 2015-10-04
上傳用戶：qunquan
這是一個(gè)C51的時(shí)鐘程序源代碼.程序用P0口輸出段碼,P1口輸出位碼,四位顯示,用PC機(jī)校時(shí).
標(biāo)簽： C51 輸出 時(shí)鐘程序 P0口
上傳時(shí)間： 2013-12-19
上傳用戶：haoxiyizhong
American Gladiator,You are consulting for a game show in which n contestants are pitted against n gladiators in order to see which contestants are the best. The game show aims to rank the contestants in order of strength this is done via a series of 1-on-1 matches between contestants and gladiators. If the contestant is stronger than the gladiator, then the contestant wins the match otherwise, the gladiator wins the match. If the contestant and gladiator have equal strength, then they are “perfect equals” and a tie is declared. We assume that each contestant is the perfect equal of exactly one gladiator, and each gladiator is the perfect equal of exactly one contestant. However, as the gladiators sometimes change from one show to another, we do not know the ordering of strength among the gladiators.
標(biāo)簽： contestants consulting Gladiator are
上傳時(shí)間： 2013-12-18
上傳用戶：windwolf2000
Preparation of a function, even to the importation of n, the function call seeking +1/n 1/2+1/4+ .... When n is odd input, the function call 1/1+1/3+ ... +1/n (using function pointer)
標(biāo)簽： function Preparation importation the
上傳時(shí)間： 2014-01-21
上傳用戶：semi1981
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