Codeforces Round #484 (Div. 2)

Codeforces Round #484 (Div. 2) B. Bus of Characters B. Bus of Characters time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output In the Bus of Characters there are nn rows of seat, each having 22

Codeforces Codeforces Round #484 (Div. 2) E. Billiard 题目连接: http://codeforces.com/contest/982/problem/E Description Consider a billiard table of rectangular size $n \times m$ with four pockets. Let's introduce a coordinate system with the origin at t

Codeforces Codeforces Round #484 (Div. 2) D. Shark 题目连接: http://codeforces.com/contest/982/problem/D Description For long time scientists study the behavior of sharks. Sharks, as many other species, alternate short movements in a certain location and

题意:给你一个序列,让你找一个k,倘若把大于等于k的元素都标记为不可用,那么剩下的所有元素形成的段的长度相同,并且使得段的数量尽量大.如果有多解,输出k尽量小的. 把元素从大到小排序插回原位置,用一个set维护前驱后继,相当于删除一个原有的段,然后将这个段切成两半,产生两个新的段.维护这次操作后所有段的长度以及各种长度的出现次数(用multiset),一旦合法,就尝试更新答案. #include<cstdio> #include<algorithm> #include<set

A. Row time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You're given a row with nn chairs. We call a seating of people "maximal" if the two following conditions hold: There are no neigh

Codeforces Round #428 (Div. 2) A    看懂题目意思就知道做了 #include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define rep(i,a,b) for (int i=a; i<=b; ++i) #define per(i,b,a) for (int i=b; i>=a; --i

题目链接:Codeforces Round #424 (Div. 2) D. Office Keys 题意: 在一条轴上有n个人,和m个钥匙,门在s位置. 现在每个人走单位距离需要单位时间. 每个钥匙只能被一个人拿. 求全部的人拿到钥匙并且走到门的最短时间. 题解: 显然没有交叉的情况,因为如果交叉的话可能不是最优解. 然后考虑dp[i][j]表示第i个人拿了第j把钥匙,然后 dp[i][j]=max(val(i,j),min(dp[i-1][i-1~j]))   val(i,j)表示第i个人拿

题目链接:Codeforces Round #424 (Div. 2) C. Jury Marks 题意: 给你一个有n个数序列,现在让你确定一个x,使得x通过挨着加这个序列的每一个数能出现所有给出的k个数. 问合法的x有多少个.题目保证这k个数完全不同. 题解: 显然,要将这n个数求一下前缀和,并且排一下序,这样,能出现的数就可以表示为x+a,x+b,x+c了. 这里 x+a,x+b,x+c是递增的.这里我把这个序列叫做A序列 然后对于给出的k个数,我们也排一下序,这里我把它叫做B序列,如果我

人生不止眼前的狗血,还有远方的狗带 A题B题一如既往的丝帛题 A题题意:询问按照12345678910111213...的顺序排列下去第n(n<=10^3)个数是多少 题解:打表,输出 1 #include<bits/stdc++.h> 2 using namespace std; 3 int dig[10],A[1005]; 4 int main(){ 5 int aa=0; 6 for(int i=1;;i++){ 7 int x=i,dd=0; 8 while(x)dig[++dd