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driblet    
n. 少许;少量;小额

少许;少量;小额

driblet
n 1: a small indefinite quantity (especially of a liquid); "he
had a drop too much to drink"; "a drop of each sample was
analyzed"; "there is not a drop of pity in that man";
"years afterward, they would pay the blood-money, driblet
by driblet"--Kipling [synonym: {drop}, {drib}, {driblet}]

Dribblet \Drib"blet\, Driblet \Drib"let\, n. [From {Dribble}.]
A small piece or part; a small sum; a small quantity of money
in making up a sum; as, the money was paid in dribblets.
[1913 Webster]

When made up in dribblets, as they could, their best
securities were at an interest of twelve per cent.
--Burke.
[1913 Webster]


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  • Persistent segment trees: Explained with SPOJ problems
    Consider the addition of this new value like an update operation on segment tree of u Then the update operation will change O(log N) nodes from last level to the 1st level There you go, the new segment tree of node c is almost same as segment tree of u, except for O(log N) nodes Hence we can only allocate memory for this new O(log N) nodes
  • What are persistent segment trees and how do I use them?
    I have come across a lot of problems which could be solve using persistent segment trees but I can't seem to understand this concept I have found some resources but I still can't understand it
  • How to free all the memory in a tree of pointers
    Well the best solution for this would be to create a persistent segment tree without pointers I can easily recreate the tree without having to delete all the memory recursively, which is very buggy and annoying to implement, especially since I'm not familiar with pointers This considerably reduces memory usage The new node looks like:
  • algorithm - Fenwick tree vs Segment tree - Stack Overflow
    A BIT requires only half as much memory as a segment tree: In cases where you have masochistic memory constraints, you are almost stuck with using a BIT; Though BIT and segment tree operations are both O(log(n)), the segment tree operations have a larger constant factor: This should not matter for most cases But once again, if you have
  • algorithm - Lazy propagation in segment tree? - Stack Overflow
    Well, I was trying to solve this Flipping coins problem on Codechef Am solving it with segment trees But getting time limit exceed I searched and found that I have to use lazy propagation But I
  • How can I implement segment trees with lazy propagation?
    Conceptually they form one binary tree The true value of a node in the binary tree is the value in stored sum tree, plus the number of leaves under the node times the sum of all carry tree values from the node back up to the root At the same time, the true value of each node in the tree is equal to the true value of each leaf node under it
  • How to implement segment trees with lazy propagation?
    First of all thanks for spending your precious time to answer my question There are few doubts in my mind which I would like to discuss First I have studied that we implement segment tree using arrays which are operated just like heap data structure Secondly,you are saying that we don't create left and right child until needed but in the link which I have given above it says that the entire
  • c++ - Dynamic version of MKTHNUM - Stack Overflow
    I am attempting to solve a dynamic version of MKTHNUM The static version involves a given array of N elements where we have to answer queries q(l, r, k) in the subarray from l to r, determining the kth element when considering the subarray sorted I have successfully addressed this initial part using a persistent segment tree
  • data structures - When to use a treap - Stack Overflow
    Treap is the blend of tree and heap The idea is to enforce BST’s constraints on the names, and heap’s constraints on the quantities Product names are treated as the keys of a binary search tree The inventory quantities, instead, are treated as priorities of a heap, so they define a partial ordering from top to bottom





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