D - Sand Fortress 思路: 二分 有以下两种构造, 分别二分取个最小. 代码: #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #defi…

Sand Fortress time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimens…

A B /* Huyyt */ #include <bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define mkp(a,b) make_pair(a,b) #define pb push_back ][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }}; using namespace std; typedef long long ll; inline v…

Description You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be described as a line of spots to pile up sand pillars. Spots are numb…

思路: 很奇怪的结论题,不好想.参考了http://codeforces.com/blog/entry/59623 实现: #include <bits/stdc++.h> using namespace std; typedef long long ll; const ll INF = 0x7f7f7f7f; ll n, H; bool check(ll m) { ll tmp = min(m, H); ) / <= n; } ll solve(ll n, ll H) { ll l =…

F - Isomorphic Strings 思路:字符串hash 对于每一个字母单独hash 对于一段区间,求出每个字母的hash值,然后排序,如果能匹配上,就说明在这段区间存在字母间的一一映射 代码: #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair…

E - Pencils and Boxes 思路: dp 先排个序,放进一个袋子里的显然是一段区间 定义状态:pos[i]表示小于等于i的可以作为(放进一个袋子里的)一段区间起点的离i最近的位置 显然,初始状态:pos[i] = 1,1 <= i <= k 状态转移: pos[i+1] = i+1 ,如果 a[i] - a[pos[i-k+1]] <= d , 因为在这种情况下pos[i-k+1] 到 i 可以放进一个袋子里,那么i+1就可以作为新的起点了 pos[i+1] = pos[…

E. Pencils and Boxes time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Mishka received a gift of multicolored pencils for his birthday! Unfortunately he lives in a monochrome world, where ev…

传送门:送你去985D: 题意: 你有n袋沙包,在第一个沙包高度不超过H的条件下,满足相邻两个沙包高度差小于等于1的条件下(注意最小一定可以为0),求最少的沙包堆数: 思路: 画成图来说,有两种可能,一种是y=h-x一次函数和常函数y=x组合,还有一种是先上升后下降的函数,注意斜率绝对值都是1: 二分答案,具体来说,我是二分了最大高度,主要是check()比较要思考,(昨天晚上没有细心写check,错过了很多AC:): 这个check中:如果最大高度 mid 比m小,说明不会有折线,直接考虑三角…

题目链接:https://codeforces.com/contest/985 ’A.Chess Placing 题意:给了一维的一个棋盘,共有n(n必为偶数)个格子.棋盘上是黑白相间的.现在棋盘上有n/2个棋子,让你全部移动到黑色格子或者白色格子,要求步数最少,并输出步数. 题解:由于黑色或者白色都是固定位置,我们对棋子位置排个序,然后全部移动到任意一个颜色,取两个最小步数的最小值就好了. #include<bits/stdc++.h> #define clr(x) memset(x,0,s…