原文链接https://www.cnblogs.com/zhouzhendong/p/AtCoder-SoundHound-Inc-Programming-Contest-2018-E.html 题目传送门 - AtCoder SoundHound Inc. Programming Contest 2018 E 题意 给定一个无向连通图,有 $n$ 个节点 $m$ 条带权边,第 $i$ 条边连接 $x_i,y_i$ ,权值为 $s_i$ ,没有重边.自环. 现在,请你给每一个节点取一个正整数点权…

A - F Time Limit: 2 sec / Memory Limit: 1024 MB Score : 100100 points Problem Statement You are given two integers aa and bb. Determine if a+b=15a+b=15 or a×b=15a×b=15 or neither holds. Note that a+b=15a+b=15 and a×b=15a×b=15 do not hold at the same…

ACM International Collegiate Programming Contest, Tishreen Collegiate Programming Contest (2018) Syria, Lattakia, Tishreen University, April, 30, 2018 Problem A. Can Shahhoud Solve it? Problem B. Defeat the Monsters Problem C. UCL Game Night Problem…

Battle Royale games are the current trend in video games and Gamers Concealed Punching Circles (GCPC) is the most popular game of them all. The game takes place in an area that, for the sake of simplicity, can be thought of as a two-dimensional plane…

A family of nn gnomes likes to line up for a group picture. Each gnome can be uniquely identified by a number 1..n1..n written on their hat. Suppose there are 55 gnomes. The gnomes could line up like so: 1, 3, 4, 2, 51,3,4,2,5. Now, an evil magician…

Consider an n \times mn×m matrix of ones and zeros. For example, this 4 \times 44×4: \displaystyle \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 \end{matrix}1000​1111​1111​1…

Consider nn initial strings of lower case letters, where no initial string is a prefix of any other initial string. Now, consider choosing kk of the strings (no string more than once), and concatenating them together. You can make this many such comp…

John loves winter. Every skiing season he goes heli-skiing with his friends. To do so, they rent a helicopter that flies them directly to any mountain in the Alps. From there they follow the picturesque slopes through the untouched snow. Of course th…

A. Cut it Out! 枚举第一刀,那么之后每切一刀都会将原问题划分成两个子问题. 考虑DP,设$f[l][r]$表示$l$点顺时针一直到$r$点还未切割的最小代价,预处理出每条边的代价转移即可. 时间复杂度$O(n^3)$. #include<cstdio> #include<cmath> #include<algorithm> using namespace std; const int N=422; const double eps=1e-8; const…

https://codeforces.com/gym/101810 A. Careful Thief time limit per test 2.5 s memory limit per test 256 MB input standard input output standard output There are consecutive buildings numbered from 1 to 109 each of which has an amount of money in it. T…

题目链接https://nanti.jisuanke.com/t/28852 题目大意是 h*w 的平面,每两个点有且仅有一条路径,类似于封闭的联通空间,然后在这h*w个点中选取(标记为1~N)N个点(给了坐标),求从1号点按1~N的顺序走到N号点的路程. 练习赛的时候没有思路,队友说可以建树,但还是不清不楚没写出来. 做法是LCA. 将封闭的联通空间建树(点的位置与字符的位置有点麻烦),然后按顺序求两点的最近的公共祖先求深度得距离,最后得路程,算是一道LCA的模板. #include <bit…

A:Zero Array 题意:两种操作, 1 p v  将第p个位置的值改成v  2  查询最少的操作数使得所有数都变为0  操作为可以从原序列中选一个非0的数使得所有非0的数减去它,并且所有数不能变为负数 思路:考虑第二种操作,显然,最少的操作数肯定是不同数的个数 用map 记录,特殊注意0的存在 #include <bits/stdc++.h> using namespace std; #define N 100010 unordered_map <int, int> mp;…

Solution A:Careful Thief 题意:给出n个区间,每个区间的每个位置的权值都是v,然后找长度为k的区间,使得这个区间的所有位置的权值加起来最大,输出最大权值, 所有区间不重叠 思路:贪心的想法,长度为k的区间的起始点肯定是某个区间的起始点,或者长度为k的区间的结束点肯定是某个区间的结束点. 因为存在最优的答案,它的起点不在某个区间的起点,那么只有两种情况. 1° 不属于任何已知区间 2° 在某个已知区间的内部 首先考虑第一种情况  如果不属于任何已知区间,那么根据贪心,我肯定…

A:Martadella Stikes Again 水. #include <bits/stdc++.h> using namespace std; #define ll long long int t; ll R, r; int main() { scanf("%d", &t); while (t--) { scanf("%lld%lld", &R, &r); if (R * R > 2ll * r * r) puts(&…

题目链接:https://codeforces.com/gym/101853 A: ✅ B: (二分图匹配) https://blog.csdn.net/qq_41997978/article/details/89165151(二分图匹配+基础数论+算术分解建图) C: (逆序对) 已过 D:✅ E:✅ https://blog.csdn.net/Link_Ray/article/details/89102023 (状压经典题) F:✅ G: 已过 https://blog.csdn.net/L…

// Coolest Ski Route #include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <utility> #include <vector> #include <map> #include <queue> #include <st…

bryce1010模板 http://codeforces.com/gym/101810 #include <bits/stdc++.h> using namespace std; #define ll long long const int MAXN=1010; int a[MAXN<<1]; int main() { int t; cin>>t; int n; while(t--) { cin>>n; for(int i=1;i<=2*n;i++)…

bryce1010模板 http://codeforces.com/gym/101810 #include<bits/stdc++.h> using namespace std; #define ll long long ll a[1000005]; int main() { int t,n,temp; ll ans,maxn,minl,x; scanf("%d",&t); while(t--) { scanf("%I64d",&n);…

bryce1010模板 http://codeforces.com/gym/101810 #include<bits/stdc++.h> using namespace std; #define ll long long const ll maxn = 1e5 + 5; const ll mod = 1e9+7; ll n; ll p[maxn]; ll vis[maxn]; int main() { ll t; scanf("%lld",&t); ll pre;…

bryce1010模板 http://codeforces.com/gym/101810 #include <bits/stdc++.h> using namespace std; #define ll long long int main() { int t; scanf("%d",&t); ll n,m; while(t--) { scanf("%lld%lld",&n,&m); ll cnt=0; if(m>n)swa…

bryce1010模板 http://codeforces.com/gym/101810 #include <bits/stdc++.h> using namespace std; #define ll long long ll lowbit(ll x) { return x&(-x); } int main() { ll t; cin>>t; ll x; while(t--) { cin>>x; ll res=1; while((x&1)==0) {…

bryce1010模板 http://codeforces.com/gym/101810 #include <bits/stdc++.h> using namespace std; #define ll long long int main() { int T; cin >> T; while (T--) { ll n, x; cin >> x >> n; if (n == 1) { cout << x << endl; contin…

bryce1010模板 http://codeforces.com/gym/101810/problem/A 大模拟,写崩了,代码借队友的...... 注意处理段与段的连接问题: #include<bits/stdc++.h> using namespace std; const int maxn=1e5+5; struct node{ long long l,r,val; node(){} node(long long ll,long long rr,long long vval) { l=…

题面 题意:T组数据,每次给你1e5个点的树(1为根),每个点有一权值,询问1-n每个节点的子树中, 至少修改几个点的权值(每次都可以任意修改),才能让子树中任意2点的距离==他们权值差的绝对值 无解输出-1 题解:画图不难发现,如果这个节点有3个儿子,也就是不包含它连向它父亲的边,它还有多于2条边的话,一定不行 因为子节点权值只能是这个节点+1或者-1,所以只能存在最多2个 那我们就又发现了,只有这个子树,可以拉成一个链的时候,才有答案, 考虑在链中的情况,如何判断修改最少的个数,使得这是个差…

题意:有\(n\)个点,\(n-1\)条边,每条边正向和反向有两个权值,且每条边最多只能走两次,有\(m\)次询问,问你从\(u\)走到\(v\)的最大权值是多少. 题解:可以先在纸上画一画,不难发现,除了从\(u\)走到\(v\)的路径上的反向权值我们取不到,其他所有边的正反权值均能取到,所以答案就是:\(sum-u->v路径的反向权值\),问题也就转换成了求\(v->u\)的权值,那么这里我们就可以用LCA来求了. 首先,令一个点为根节点,然后求出\(v\)到根节点的距离和根节点到\(u\…

B Breaking Branches 题意:两个人比赛折枝,谁剩下最后1,无法折出整数即为输 思路:树枝长n,若是奇数,则Bob胜出,若是偶数,则Alice胜出,且需要输出1: 1 #include<stdio.h> 2 int main(){ 3 int n; 4 scanf("%d",&n); 5 if(n%2==0){ 6 printf("Alice\n"); 7 printf("1\n"); 8 }else{ 9 p…

2018 ACM-ICPC, Syrian Collegiate Programming Contest A Hello SCPC 2018! 水题 B Binary Hamming 水题 C Portals 思路:并查集维护连通性 代码: //#pragma GCC optimize(3) //#pragma comment(linker, "/STACK:102400000,102400000") //c++ // #pragma GCC diagnostic error &quo…

2018 German Collegiate Programming Contest (GCPC 18) Attack on Alpha-Zet 建树,求lca 代码: #include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bit…

题目来源 The 2018 ACM-ICPC China JiangSu Provincial Programming Contest 35.4% 1000ms 65536K Persona5 Persona5 is a famous video game. In the game, you are going to build relationship with your friends. You have N friends and each friends have his upper b…

2018 AICCSA Programming Contest A Tree Game B Rectangles 思路:如果存在大于0的交面积的话, 那么肯定能找到一条水平的直线 和 一条垂直的直线, 使得水平直线的左右两边点的个数相等且为n, 垂直直线的左右两边点的个数相等且为n 也就是说不能有点在这两条线上, 否则交面积为0 然后左上角的点和右下角的点配对, 左下角的点和右上角的点配对 代码: #pragma GCC optimize(2) #pragma GCC optimize(3) #…