Given an array A of N length and you are also given number B. You need to find the number of subarrays in A having sum less than B. We may assume that there is no overflow. Problem Constraints 1 <= N <= 10000, -100<=A[i]<=100, 0<=B<=10000 Example: INPUT:: A---> [1...
C. Increase Subarray Sums 传送门:Problem - 1644C - Codeforces 题目大意就是:给你一个序列,算出序列和为sum,求是否有连续区间和大于等于sum,有就输出NO,否则YES,并且连续区间不能是[1,n],也就是整个序列,我的思路就是构建两个DP数组,既然区间不能是1->n,那么一个dp从1->n-1,第二个从2->n,求dp...
Consider the first sample test. The optimal solution is obtained if the first subarray contains the first element only, the second subarray spans the next three elements and the last subarray contains the last element only. The sums of these subarrays are5,9and1, correspondingly. Consider the ...
Educational Codeforces Round 123 (Rated for Div. 2) C. Increase Subarray Sums 传送门:Problem - C - Codeforces 题目大意就是:给你一个数组a,长度为n,再给你一个数为k,定义ans数组,求ans[i]的每个元素,i从0到n.ans[i]代表给数组中i个元素加上k后的最大连续子区间和。最大子区间长度可以是0,也...
Hello Codeforces. Recently I faced a problem which I couldn't solve in an hour. It is as follows: Max Sum Subarray of atleast 2 numbers. Of course, for just max sum subarray it is Kadane's algorithm in O(n) time, however, I couldn't think of a way to solve for atleast 2 numb...
Hi there, Codeforces community! This question has been bugging me for a while now, so I thought I would share it with you, and see if we can work out the answer together. The problem is the following:given an array ofninteger numbers (positive and negative), find a (contiguous) subarr...
The same will apply for bigger sizes as well. Columniwill be equal to columni - 1but with range[i, N - i + 1]increased by 1. This can be achieved in linear time with prefix sums. LetandFibe the contribution of elementAito the final answer (divided byAi). Initially...
Convert the statement to it's equivalent in terms of prefix sums i.eprer−prel<=kprer−prel<=k. Now iterating overrr, we need to find number ofllsuch thatprel>=prer−kprel>=prer−k, which can be computed using ordered set of allprelprel. ...
publicintshortestSubarray(int[]A,intK){intptr=0;intsum=0;intsol=Integer.MAX_VALUE;for(inti=0;i<A.length;++i){sum+=A[i];while(sum>=K){sol=Math.min(sol,i-ptr+1);sum-=A[ptr++];}}return(sol==Integer.MAX_VALUE)?-1:sol;} ...
sub_index = i - k ksum = ksum + array[add_index] - array[sub_index] max_ksum = math.max(ksum, max_ksum) end return max_ksum end for k = 1, n do print(k, maxArray(k)) end Is there any algorithm with lower time complexity? For example, O(N log N) + additional memor...