[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"$fS2-_IHE1T2QHTs3uC5dZmszODUJSU0IMCplgldncGdo":3},{"answer":4,"createTime":5,"id":6,"options":7,"origin":12,"question":19,"related":20,"source":30,"type":31},[],"2024-03-31 21:37:18",136454445,[8,9,10,11],"T(n)=3nlog2n","T(n)= T(n-1)+1,T(1)=1","T(n)= 2n2","T(n)= T(n\u002F2)+1, T(1)=1",{"count":13,"courseId":14,"courseImg":15,"courseName":16,"workId":17,"workName":18},15,"eec0cd1f03c083aa5024bde75bf2cf06","https:\u002F\u002Ftihai-oss-cloud.itihey.com\u002Fimg\u002F7e2d5ad53be3d12aacc56ee37db405b7.jpg","算法分析与设计（22级）","work_33191362","第三章 递归与分治策略 测试题","T(n)表示当输入规模为n时的算法效率,以下算法效率最优的是( )",[21,32,41,50,59,68,77,86,89,98],{"answer":22,"createTime":5,"id":23,"options":24,"question":29,"source":30,"type":31},[],136454438,[25,26,27,28],"O(n2)","O(nlogn)","O(logn)","O(n)","分治法解选择问题的最好效率属于( )","v1",0,{"answer":33,"createTime":5,"id":34,"options":35,"question":40,"source":30,"type":31},[],136454439,[36,37,38,39],"子问题的解可以合并","子问题不能够重复","子问题必须是一样的","原问题和子问题使用相同的方法解","使用分治法求解不需要满足的条件是()",{"answer":42,"createTime":5,"id":43,"options":44,"question":49,"source":30,"type":31},[],136454440,[45,46,47,48],"int binarySearch(int a[], int n, int x){ if(n &gt; 0 &amp;&amp; x &gt;= a[0]){ int low = 0, high = n-1;while(low &lt; high){ int mid=(low+high+1)\u002F2;if(x &lt; a[mid])high=mid-1;else low=mid;}if(x==a[low]) return low;}return &ndash;1;}","int binarySearch (int a[], int n, int x){ int low=0, high=n-1;while(low&lt;high-1){ int mid=(low+high)\u002F2;if(x&lt;a[mid])high=mid;else low=mid;}if(x==a[low]) return low;else return &ndash;1;}","int binarySearch(int a[], int n, int x){ int low=0, high=n-1;while(low+1!=high){ int mid=(low+high)\u002F2;if(x&gt;=a[mid]) low=mid;else high=mid;}if(x==a[low]) return low;else return &ndash;1;}","int binarySearch(int a[], int n, int x){ int low=0, high=n-1;while(low&lt;=high){ int mid=(low+high)\u002F2;if(x==a[mid]) return mid;if(x&gt;a[mid]) low=mid;else high=mid;}return &ndash;1;}","对于下列二分查找算法,以下正确的是( )",{"answer":51,"createTime":5,"id":52,"options":53,"question":58,"source":30,"type":31},[],136454441,[54,55,56,57],"棋盘覆盖问题","选择问题","0\u002F1背包问题","归并问题","以下不可以使用分治法求解的是( )",{"answer":60,"createTime":5,"id":61,"options":62,"question":67,"source":30,"type":31},[],136454442,[63,64,65,66],"O(logn)和O(n2)","O(n)和O(n2)","O(nlogn)和O(n)","O(nlogn)和O(n2)","快速排序算法在最好情况下和最坏情况下的时间复杂度分别为 ( )",{"answer":69,"createTime":5,"id":70,"options":71,"question":76,"source":30,"type":31},[],136454443,[72,73,74,75],"问题规模相同,问题性质相同","问题规模相同,问题性质不同","问题规模不同,问题性质相同","问题规模不同,问题性质不同","分治法的设计思想是将一个难以直接解决的大问题分割成规模较小的子问题,分别解决子问题,最后将子问题的解组合起来形成原问题的解.这要求原问题和子问题( )",{"answer":78,"createTime":5,"id":79,"options":80,"question":85,"source":30,"type":31},[],136454444,[81,82,83,84],"用中位数的中位数方法寻找划分基准","取子序列的第一个元素作为划分基准","随机选择一个元素作为划分基准","以上皆可行.但不同方法,算法复杂度上界可能不同","在寻找n个元素中第k小元素问题中,如快速排序算法思想,运用分治算法对n个元素进行划分,如何选择划分基准?下面( )答案解释最合理",{"answer":87,"createTime":5,"id":6,"options":88,"question":19,"source":30,"type":31},[],[8,9,10,11],{"answer":90,"createTime":5,"id":91,"options":92,"question":97,"source":30,"type":31},[],136454446,[93,94,95,96],"(4K-1)\u002F2","(2K-1)\u002F3","(2K-1)\u002F2","(4K-1)\u002F3","在任何一个2k*2k的棋盘覆盖中,用到的L型骨牌个数恰好为( )",{"answer":99,"createTime":5,"id":100,"options":101,"question":106,"source":30,"type":31},[],136454447,[102,103,104,105],"无法判断","最差效率","最优效率","两者都不是","快速排序时,如果所有元素都相等,这时取得的效率属于( )"]