consecutive even integers calculator

Title Description:http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1138A positive integer n is given, and N is written as a number of consecutive numbers and forms (length >= 2). For example, n = 15, can be written as 1 + 2 + 3 + 4 + 5, can also be written as 4 + 5 + 6, or 7 + 8. If it cannot be written as a number of consecutive integers, the output

Describe:Implement input a set of integers greater than or equal to 0, according to the order from small to large output, after sorting there are consecutive numbers, only the minimum and maximum number of consecutive number of two.Input:A set of integers greater than or equal to 0, regardless of illegal input, separat

("%d", n) = =1 n>0) { for(i=0; i"%d", nums[i]); } for(res=prev=sum=-1, first=nums[0],last=nums[n-1], i=0; iif(prev0) {if(nums[i]>=0) {//prev Start increasing, update candidate of first-TMPTmp=prev=nums[i];//Update candidate of result--sum if(prev>sum) {sum=prev; } } }Else{Prev+=nums[i];//Prev Stop increasing, update first, last, Res if(nums[i]0) {if(sum>res) {res=sum; first=tmp; last=nums[i-1]; } }//Update candidate of result--sum

See this problem in the group, do it with python,defFind (array): v_sum= Greatest =0 forAinchArray:v_sum+=a v_sum= 0ifV_sum ElseV_sum Greatest= V_sumifV_sum > GreatestElseGreatestifV_sum = =0:greatest=array[0] forAincharray:greatest= AifGreatest ElseGreatestPrintgreatestarr1= [1, 2,-4, 3, 3]find (arr1)Idea: from left to right sliding scale, if the v_sum is positive, you can continue to add to the back of the element,If the v_sum has been negative, it is necessary to clear the original v_sum to 0

Title Link: http://lightoj.com/volume_showproblem.php?problem=1278Test instructions: give you a number n (nFor example: 15 = 7+8 = 4+5+6 = 1+2+3+4+5, so 15 corresponding answer is 3, there are three kinds;We are now equivalent to known arithmetic progression and sum = N, the other first item is A1, Total m, then am = a1+m-1;sum = m* (a1+a1+m-1)/2-----> a1 = sum/m-(m-1)/2A1 and M must be integers, so sum%m = 0 and (m-1)%2=0, so M is the factor of sum,

Topic Link: PortalTest instructions: Divides the given n into successive numbers and has at least two numbers to see how many options are available.Analysis:A + (A + 1) + (A + 2) + ... +(A + k-1) = N; ===> (2*a + k-1) * k = 2*n;===> (2*a-1) *k = 2*n-k*k;===> 2*a-1 = 2*n/k-K;The left side of the equation is odd, then the right side must be odd, then K must be an odd number of nSo the n factor decomposition is possible.The code is as follows:#include Light OJ 1278 Sum of