When I was sovling the[problem:356C]，my java code got TEL but c++ code got AC,why? The core code of them is the same. JAVA:4857792 C++:4857784

When I solve 313C - Илья и матрица, I got WA on testcases 11 :(.my submission is 3805151. Can you help me ? :)

I saw many people got wrong on test10 in problem A #174 DIV1 . It says "wrong answer 103366th numbers differ — expected: '-1310.8301270', found: '-1310.8321289', error = '0.0000015",what's wrong? 3352253

I advice we open a new webpage in codeforces and collect all the editorials.

I am a student in China. I used to practice in POJ,ZOJ,or HDUoj,but when I came across a problem I can't get much help especially when I can't understand the tuturial. But at codeforces I can have a heated discuss with others. I really get much help and I thanks all of you.

I still can't figure out how to solve this problem after I read the tuturial. Can you write your idea carefully :(

what's wrong with my code of 158_DIV2_C? Here is my submission:2847080

My rating is just 1453 but I still wanner write my own tutorial to share with you.

A. 252A - Маленький xor This is a simple problem. We can write a algorithm using brute force. The complexity of it is O(n^3). My code 2713341.

B.252B - Разупорядочивание массива This is a greedy problem. There are several special cases: 1. There is only one or two interger. Answer is "-1". 2. There are three interger and they are placed like "1 2 1" or "2 1 2", whatever you do you will make an sorted array. 3. All the numbers are the same so we can't make any operation. 4. Here are other cases: We find the first unequal pair searching from the head of the array, then we swap them. We will get the unsorted array unless what happen below: The pair are the first two numbers and the second is the max number or the min number. So we just search the unequal pair from the trail and swap them then we will get the right answer.The complexity of it is O(n). My code 2714096

C.251A - Точки на прямой This is a DP problem. We user two pointers and one points to the left most interger and the other points to the right most interger.For every left points we need find out the right most interger, then we just calculate C(r-l+1,2) means we pick up two intergers from the sequences freely. Because the position of the right interger can't decrease so the complexity of it is O(n*2).My code 2704914

These are the problem I solved in the contest. Sorry for my poor English and bad code ability. You may seek the official editorial for more infomation.

As a newbie,I really wanner learn from the problem.But some are too hard to solve even after the contest.So can we make a displine that every auther who hold the contest should write the editorial or find someone else to write it? If so, it will help us newbies a lot.