感觉这次的题目顺序很不合理啊...
A. Combination Lock
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he‘s earned fair and square, he has to open the lock.
The combination lock is represented by n rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
直接模拟就好了...每一位上两个数的环上距离
B. School Marks
Little Vova studies programming in an elite school. Vova and his classmates are supposed to write n progress tests, for each test they will get a mark from 1 to p. Vova is very smart and he can write every test for any mark, but he doesn‘t want to stand out from the crowd too much. If the sum of his marks for all tests exceeds value x, then his classmates notice how smart he is and start distracting him asking to let them copy his homework. And if the median of his marks will be lower than y points (the definition of a median is given in the notes), then his mom will decide that he gets too many bad marks and forbid him to play computer games.
Vova has already wrote k tests and got marks a1, ..., ak. He doesn‘t want to get into the first or the second situation described above and now he needs to determine which marks he needs to get for the remaining tests. Help him do that.
这道题的简单思路就是在中位数满足条件的情况下
让整个序列都尽量小，以尽量满足第一个条件
首先如果读入的中位数下限是x，那么显然最终序列的中位数也是x
我们假设最终中位数不是x，而是原序列中第i小的数a[i]
那么a[i]以左不够n/2的数用1填，a[i]以右不够n/2的数用a[i]填
然而这个时候如果我们用x作为中位数，x的位置只会小于等于i
先看小于的时候，左边填的1的个数更多了，右边填的x的个数更少了，显然总和会小
等于的时候，左边填的1不变，右边填的个数也不变，只是用x填，而x<=a[i]，所以总和会更小
因为题目保证k<n，总要填一个数，所以不论如何中位数取x总没错
需要判断的就是左右的个数和总和，但是要注意一些细节
当然可能是我的方法不大好所以产生了很多细节需要判断
类似于：原序列中没有数、原序列中没有大于等于x的数...
C. Ice Cave
You play a computer game. Your character stands on some level of a multilevel ice cave. In order to move on forward, you need to descend one level lower and the only way to do this is to fall through the ice.
The level of the cave where you are is a rectangular square grid of n rows and m columns. Each cell consists either from intact or from cracked ice. From each cell you can move to cells that are side-adjacent with yours (due to some limitations of the game engine you cannot make jumps on the same place, i.e. jump from a cell to itself). If you move to the cell with cracked ice, then your character falls down through it and if you move to the cell with intact ice, then the ice on this cell becomes cracked.
Let‘s number the rows with integers from 1 to n from top to bottom and the columns with integers from 1 to m from left to right. Let‘s denote a cell on the intersection of the r-th row and the c-th column as (r, c).
You are staying in the cell (r1, c1) and this cell is cracked because you‘ve just fallen here from a higher level. You need to fall down through the cell (r2, c2) since the exit to the next level is there. Can you do this?
这道题的样例好讨厌啊><
绕了一大圈然后就决定用网络流做了...虽然觉得好不科学啊cf上的题居然还有那么长的代码？
最后光荣RE 也不打算去找哪里出错了...毕竟是一道不那么难的题
对于一个未破裂的点，我们有时需要找到一条全是未破裂格子的路绕回到这个点
其实非常简单，只要它的四周存在一个未破裂的点，走到那里，然后再回来就可以了...
其他情况完全没必要去考虑
这样了之后，整道题的思路就简单了好多
首先考虑起点和终点重合的情况，只要四周有一个未破裂的点就可以了
然后考虑其他情况，首先我们需要起点到终点存在一条未破裂格子组成的道路
当然如果两点相邻那就不用考虑了
然后分情况讨论
如果终点是破裂的话，那就直接YES
如果未破裂，意味着要从这个点的四周再找一个未破裂的格子
所以只要存在一条起点到终点的路径，并且终点的四周存在>=2个未破裂的格子就可以了
D. Bad Luck Island
The Bad Luck Island is inhabited by three kinds of species: r rocks, s scissors and p papers. At some moments of time two random individuals meet (all pairs of individuals can meet equiprobably), and if they belong to different species, then one individual kills the other one: a rock kills scissors, scissors kill paper, and paper kills a rock. Your task is to determine for each species what is the probability that this species will be the only one to inhabit this island after a long enough period of time.
看到这么简单的D的时候简直惊呆了...
定义一个f[i,j,k]表示有i个石头j个剪刀k个布的概率
初始情况f[r,s,p]=1，然后从前往后推比较方便
当石头减少1个的时候，意味着一个石头和一个布见面，概率是i*k/(i*k+i*j+k*k)
注意这里的分母不是C(i+j+k,2),因为当两个同种类的村民见面的时候不会发生任何事情可以忽略
另外两种情况同理
E. Infinite Inversions
There is an infinite sequence consisting of all positive integers in the increasing order: p = {1, 2, 3, ...}. We performed n swapoperations with this sequence. A swap(a, b) is an operation of swapping the elements of the sequence on positions a and b. Your task is to find the number of inversions in the resulting sequence, i.e. the number of such index pairs (i, j), that i < j and pi > pj.
看到这么简单的E的时候简直惊呆了...
离散一下数据然后模拟一下然后统计一下逆序对好像就可以了...
当然这只是一部分的答案
还有一部分是没有在读入中出现的那些数字
对于每一个再读入里出现的数字，只要统计那些在它原位与当前位置之前的没有出现过的数就可以了
这个可以预处理，然后前缀和，最后统计一下就好了
（看到一个TAG是trees，表示非常不理解...
要加油啊....
另外，祝
策爷 松爷 约大爷 数国队 杜教
CTSC成功虐场
2/.May