find minimum mathematica

Label: Ar SP problem BS nbsp ip c array first I did two questions today. I didn't solve the second question. I found it too troublesome. I gave up ...... Do it tomorrow when your mind is clear. The first question is the question in the title. Find the smallest value in a rotating sorting array. There are two different requirements for this problem, that is, the duplicate element and the duplicate element are not considered. My solution is to save

C language: enter 10 integers, and the program will find the maximum and minimum values.Program: # Include Result: enter 10 integers: 1 2 3 56 23 6 767 32 11 567max = 767 min = 1. Press any key to continue...

var arr = [1,5,2,56,12,34,21,3,5]math.min.apply ({},arr) // 1 Math.max.apply ({},arr) // arr.sort ((m,n)=>m-n)/ / [1, 2, 3, 5, 5, a. Arr.sort ((m,n) =>n-m) //[56, 34, 21, 12, 5, 5, 3, 2, 1]To repeatvar arr = [2,1,4,3,2,4,2,3,4,2,6,5,5]var obj = {}; var arrnew = []; for (var i=arr.length-1;i>=0;i--) { if(! Obj[arr[i]]) { true; Arrnew.push (Arr[i]);} } Console.log (arrnew,obj)// [5, 6, 2, 4, 3, 1]Array.prototype.unique2 =function(){ This. sort ();//Sort Fir

[Cpp]# Include # Include # Include /*************************************** **************************************** **************************************** ****************************************** ** Begin: 2012 10 12* *** Number of replies* *** Find the minimum number of input files greater than a single number* *** End: 2012 10 12******************************************** ***************************

C ++, Kruskal cruise Karl algorithm to find the Minimum Spanning Tree, kruskal cruise The first blog. The principle of finding the smallest Spanning Tree in kackar: first, we can look at n points as n Independent Sets and arrange all sides quickly (from small to large ). Then, enumerate each edge in the arranged order to determine whether the two points connected by this edge belong to a set. If yes, the ed

LongMod_mul (Long LongALong LongBLong LongMoD) { Long Longsum=0; while(b) {if(b1) sum= (sum+a)%MoD; b>>=1; A= (a+a)%MoD; } returnsum;}//A^b%mod Fast PowerLong LongQuk_mul (Long LongALong LongBLong LongMoD) { Long Longqsum=1; while(b) {if(b1) qsum=Mod_mul (qsum,a,mod);//mod is not a very big time. This step can remove the B>>=1; A=Mod_mul (A, A, Mod);/}returnqsum;}Long LongGetjie (Long LongALong LongM) { Long LongD =gcd (a,m); if(d! =1)return-1; Long Longm=Phi (M); //and then split

Topic: Sticks Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others) Total submission (s): 364 Accepted Submission (s): 116 Problem Descriptiongeorge took sticks of the same length and cut them randomly until all parts became at most units lon G. Now he wants to return sticks to the original state, but he forgot what many sticks he had originally and how long they were originally. Him and design a prog

Test instructions: Finding the smallest element from a translated arrayIdea 1: The simplest idea is to sort and then take the first element, the complexity O (N*log (N))Code Listing 1:public class Solution {public int findmin (int[] num) { arrays.sort (num); return num[0];} }Idea 2: Traverse, compare the first of the two orderby ordinal numbers, 4 5 6 7 0 1 2 complexity O (N)Code Listing 2: public int findmin (int[] num) {//solution 1 times int min = num[0], currmin

Turn from http://www.cnblogs.com/tiandsp/archive/2012/01/07/2316006.html thanks to DSP TianClearCLCimg = imread (' 124.jpg '); % Read image from graphics file[m n T] = size (IMG); % Get pattern sizemax = 0;min = 256;Avg = 0;For I =1:1:mfor j = 1:1:nIf IMG (i,j) Min = img (i,j);EndIf IMG (i,j) >maxMax = img (i,j);EndAVG = AVG +double (img (i, j));EndEndAvg = uint8 (avg/(m*n));MNTMaxMinAvgHow to read and write image files in MATLAB and simple processing of imagesThe MATLAB image processing Toolbox

A simple two-point search, it should be noted that the two-point search, left=mid+1, right=mid-1;In this question, it is necessary to pay special attention to the judgment statement:if (Num[len-1]left=mid+1;classSolution { Public: intFindmin (vectorint> num) { intlen=num.size (); intLeft,right,mid; Left=0; Right=len-1; while(leftRight ) {Mid=left+ (Right-left)/2; if(num[len-1]Num[mid]) Left=mid+1; Else Right=mid-1; } returnNum[left]; }};Fin