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:(Define (N-RT x n try-average-time) (Fixed-point (Repeat Average-damp try-average-time) (Lambda (y) (/x (Fast-expt y (-N 1))))))This allows you to arbitrarily specify the n-th equation and the corresponding average damping times. Start with the 5-time equation and see if the test results are consistent with my predictions.The test found that my push was too unreliable. It is found that the 4,5,6,7 quadratic equation can be convergent by 2-times average damping.Continue to guess (

bzoj3381[usaco2004 Open]cave Cows 2 The two cows in the caveTest instructionsRMQ problem. Sequence length ≤25000, number of questions ≤25000.ExercisesDouble.Code:1#include 2#include 3#include 4 #defineInc (I,J,K) for (int i=j;i5 #defineMAXN 251006 using namespacestd;7 8InlineintRead () {9 CharCh=getchar ();intf=1, x=0;Ten while(ch'0'|| Ch>'9'){if(ch=='-') f=-1; Ch=GetChar ();} One while(ch>='0'ch'9') x=x*Ten+ch-'0', ch=GetChar (); A returnf*x; - } - intn,m,

need to test it several times. To facilitate the test, I wrote the following method: (define (n-rt x n try-average-time) (fixed-point ((repeat average-damp try-average-time) (lambda (y) (/ x (fast-expt y (- n 1)) ) )) 1.0)) In this way, you can randomly specify the n and the corresponding average damping times, and test from five equations to see if the test results match my guess. The test found that my guess is too unreliable. The test found that the, and equations can all achieve c

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emphasized.Quick sort speeds up in practice in the following ways: Insert sort for sequences with small dividers Get median elements from a sample A quick sort is sensitive to repeating element values, when too many repeating elements are nearly two times the sort process, and the merge sort is relatively insensitive.The method of dealing with repeating elements is to use the 3-point method (3-way partitioning), that is, select the element a[j], the array is divided into less

)/* ------------------------------------------------------------------------------------------------------------- --------------------*/Proof: In the worst case, the second small element in n elements can be found using n + [Lg[n]]-2 comparisons. (Hint: Find the smallest element at the same time)General search: First find the minimum value, using a n-1 comparison, and then find the second small value, using the n-2 times comparison. A total of 2n-3 ti

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