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There are m available work queues and n jobs (m >= n > 0). T

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There are m available work queues and n jobs (m >= n > 0). The jobs are independent of each other. A job can be executed on any work queue at any time, but once started, it will fully occupy the work queue until finish. For job denoted as i (1 <= i <= n), it will take t_i time to run on a work queue, and it has to be executed before the deadline d_i, or it is deemed abandoned. Please design an efficient algorithm that tries to maximize the number of jobs that can be arranged to execute before their respective deadlines. If possible, please also describe a scenario that the algorithm you propose might fail, a concrete example is preferred.
There are m available work queues and n jobs (m >= n > 0). T
现有m个工作队列和n个作业（m>=N> 0）.这些作业是相互独立的.一个作业可以在任何时候在工作队列中工作,不过一旦开始,就完全占据该工作队列直到完成该作业为止.一个作业i（1
填空 1.There is _______(一个）"m" and ________"n" in the work "ma 已知M+｛1,3,t｝N=｛t-t+1｝若M∪N=M n=m/M 若m×m+n×n-6n+4m+13=0,则m×m-n×n=? 2x+n/m=m/n-(m/n+n/m)x（m+n不等于0） 设A是m*n矩阵,m>n,证明|AA^T|=0 lt's time for Meimei and ---to work.a.Yes,l'm writingb.l'm n 规定:区间【m,n】的长度为n-m(n大于m).设集合A=【0,t】(t大于0) 已知:2m-5n=0 求下式的值 （1+n/m-m/m-n）/(1+n/m-m/m-n) 已知2m-5n=0,求值:(1+n/m-m/m-n)÷(1-n/m-m/m+n) 数论:已知m>0,n>0,(m,n)=1,求证方程x^m=y^n的全部整数解可以由x=t^n,y=t^m给出,其中t取任 m=(m%n==0)?n: