}intmain(){intx, n;intc =0; cin>>n;for(intk=2; k<=2*n; ++k){for(intkk=1; kk<k; ++kk){ cin>>x; a[c].x=x; a[c].i=k; a[c++].j =kk; } } sort(a, a+c, cmp);intcnt =0;for(inti=0; i<c; ++i){if(!b[a[i].i] && !b[a[i].j]){ b[a[i].i]...
#include <bits/stdc++.h> using namespace std; #define int long long #define ios ios::sync_with_stdio(false), cin.tie(0), cout.tie(0) vector<int> ve(33); void init() { ve[1] = 1; ve[2] = 1; for (int i = 3; i <= 30; i++) { ve[i] = ve[i - 1] + ve[i...
The first line of the input contains a single integern(2 ≤ n ≤ 100 000) — the number of strings. The second line containsnintegersci(0 ≤ ci ≤ 109), thei-th of them is equal to the amount of energy Vasiliy has to spent in order to reverse thei-th strin...
Since I am currently focusing on DIV2 A,B,C,D only, please provide me the exhaustive list of topics for problems till DIV 2D. Since many of you have a lot of experience so please guide me with what you observed. Thank you.
而选择了第 i 个位置就意味着上一个选择的位置 j 满足a_{j+1}+a_{j+2}+\ldots+a_{i-1}<=x 。设 p 是满足条件的最小的 j 。因此, dp[i]=a[i]+\min_{j=p}^{i-1}\{dp[j]\} . p 可以采取滑动窗口的方式求得(deque不断往尾巴处放入 a_{i-1},当deque内sum大于 x 时,删除队首...
codeforces #260 DIV 2 C题Boredom(DP) 题目地址:http://codeforces.com/contest/456/problem/C 脑残了。。DP只DP到了n。。应该DP到10w+的。。 代码如下: AI检测代码解析 #include <iostream> #include <cstdio> #include <string> #include <cstring>...
题目地址:http://codeforces.com/contest/460/problem/C 这个题是用二分枚举最小值,然后判断能否在规定的次数内使得所有的数都达到这个值。判断的时候要用贪心的方法判断,从左往右遍历,这时候需要让每次浇花的范围尽量向右。所以当到达一个不得不浇花的地方时,要继续占用所需要的浇花次数。当浇花次数不够用的时候,...
great contest and Very interesting problemset :) ,but HUGE skill gap between div2B and div2C → Reply » LMydd0225 21 month(s) ago, # | ← Rev. 2 +52 Interesting Problems and quick editorials :3 I like problems 1A 1C very much, but have not enough time for 1D... Also...
Codeforces Round 915 (Div. 2) (A~E) 题解写得有点晚了,今天白天有点摸鱼,果咩qwqA.Constructive Problems显然斜着放是最优的,这样的话每一行每一列都至少需要有一个,答案即是 。 B.Begginer's Zelda显然我们每一次肯定是选择两个叶子节点来进行操作是最优的,这样一来,每一次操作都会消除掉两个叶子节点...
(x)&(-x) using namespace std; const int N=2e3+10; void solve() { int n; cin>>n; vector<int>a(n); int cnt0=0,cnt1=0,cnt2=0; vector<int>b; map<int,int>mp; for(int &x:a){ cin>>x; if(x>0)cnt1++;//统计正负数 else if(x<0)cnt2++; else cnt0++; if(x!=0...