Nature_Ran的专栏

最少拦截系统 点我找原题此题的贪心法则是这样的，先用一台系统拦截，直到无法拦截下一颗导弹为止，再开另一台系统，这时拦截下一颗导弹时，要比较这颗导弹的高度离哪台系统的高度最近就用哪台系统拦截，若都拦截不了，就再开一台系统，就这样可以算出最少的系统数量。这里要注意（也是我一直所忽略的地方）就是同一台系统可以拦截到高度相同的两颗或多颗导弹。Time Limit : 2000/1000ms (Java/

FatMouse' Trade 点我找原题题目大意：老鼠想用猫粮去换猫的豆子，但是猫有若干个仓库，每个仓库有J【i】磅的豆子，需要F【i】磅的猫粮，而每对于每个仓库而言，豆子和猫粮是按比例交换的，就是说J【i】*a%磅的豆子换F【i】*a%磅的猫粮，所以对于贪心法则，只要先选择J【i】/F【i】的值最大的换就能换到最多的豆子。Time Limit : 2000/1000ms (Java/Othe

Moving Tables 点我找原题这题只要把所有的偶数全部换成其对面的奇数房间数，就可以累计每个奇数房间所经过的次数，找到最大的那个数，就可以求出最少的时间。Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 26   Accepted

Function Run Fun 点我找原题Time Limit : 1000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 33   Accepted Submission(s) : 12Font: Times New Roman | Verdana | Georg

Calling Extraterrestrial Intelligence AgainTime Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 5   Accepted Submission(s) : 5Font: Times New Roman | V

Jugs点我找原题Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special JudgeIn the movie "Die Hard 3", Bruce Willis and Samuel L. Jackson were confronted with the following puzzle. They were given a

FatMouse' TradeTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 53691    Accepted Submission(s): 17943Problem DescriptionFatMouse prepared M poun

HikingTime Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 483    Accepted Submission(s): 259Special JudgeProblem DescriptionThere are n soda conve