题目:Click here #include <bits/stdc++.h> using namespace std; typedef long long ll; const int INF = 0x3f3f3f3f; +; int n, m; ]; ]; int main() { while( ~scanf("%d%d", &n, &m ) ) { memset( b, , sizeof(b) ); ; i<=m; i++ ) { ; j<=n…

C. Replacement Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/570/problem/C Description Daniel has a string s, consisting of lowercase English letters and period signs (characters '.'). Let's define the operation of replac…

C. ReplacementTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/570/problem/C Description Daniel has a string s, consisting of lowercase English letters and period signs (characters '.'). Let's define the operation of replace…

A. Elections time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The country of Byalechinsk is running elections involving n candidates. The country consists of m cities. We know how many peopl…

Description The country of Byalechinsk is running elections involving n candidates. The country consists of m cities. We know how many people in each city voted for each candidate. The electoral system in the country is pretty unusual. At the first s…

A. Elections time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The country of Byalechinsk is running elections involving n candidates. The country consists of m cities. We know how many peopl…

A - Elections 题意: 每一场城市选举的结果,第一关键字是票数(降序),第二关键字是序号(升序),第一位获得胜利. 最后的选举结果,第一关键字是获胜城市数(降序),第二关键字是序号(升序),第一位获得胜利. 求最后选举获胜者. 思路: 直接模拟就可以. 代码: /* * @author FreeWifi_novicer * language : C++/C */ #include<cstdio> #include<iostream> #include<cstrin…

题意:给定一个字符串,里面有各种小写字母和' . ' ,无论是什么字母,都是一样的,假设遇到' . . ' ,就要合并成一个' .',有m个询问,每次都在字符串某个位置上将原来的字符改成题目给的字符,问每次须要多少次合并次数.才使字符串没有' .. ' 思路:最原始的想法,就是对于每一次询问,都遍历整个字符串.这样时间复杂度o(n*m),就高达10^10方,非常明显会tle. 换下思路,事实上每次询问所改变的字符都会保留到下一次.也就是下一次的次数就会受到上一次的影响,那么我仅仅要就算出第一次的…

思路:把n分成[1,n/2],[n/2+1,n],假设m在左区间.a=m+1,假设m在右区间,a=m-1.可是我居然忘了处理1,1这个特殊数据.被人hack了. 总结:下次一定要注意了,提交前一定要看下边界数据,不要急着交. 题目链接:http://codeforces.com/problemset/problem/570/B <pre name="code" class="cpp">#include<bits/stdc++.h> using…

A. Elections time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The country of Byalechinsk is running elections involving n candidates. The country consists of m cities. We know how many peopl…

题:https://codeforces.com/contest/570/problem/D 题意:给定一个以11为根的n个节点的树,每个点上有一个字母(a~z),每个点的深度定义为该节点到11号节点路径上的点数.每次询问a,ba,b查询以aa为根的子树内深度为bb的节点上的字母重新排列之后是否能构成回文串.分析:很明显是个树上启发式合并.显然,只要深度为bb结点的所有颜色中,至多有一种的数量为奇数就可以构成回文串了. #include<bits/stdc++.h> using namespa…

D. Tree Requests time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Roman planted a tree consisting of n vertices. Each vertex contains a lowercase English letter. Vertex 1 is the root of the…

Problem D: 题意:给定一棵n个点树,每个点有一个字母,有m个询问,每次询问某个节点x的子树中所有深度为k的点能否组成一个回文串 分析:一堆点能组成回文串当且仅当数量为奇数的字母不多于1个,显然这个状态可以用二进制表示 那么对于单个询问如何快速找到所有符合要求的点呢? 这里可以考虑树的dfs序,我们把深度相同的点按照dfs序加入集合中,易知在同一颗子树中的点肯定形成了一个连续的区间. 因此每次可以通过二分子树根节点的进入dfs序的时间和出dfs序的时间来找到这个区间 找到区间后可以根据预…

题目:Click here 题意:看一下题目下面的Note就会明白的. 分析:一开始想的麻烦了,用了树状数组(第一次用)优化,可惜没用. 直接判断: #include <bits/stdc++.h> using namespace std; typedef long long ll; const int INF = 0x3f3f3f3f; ; int n, m; char str[M]; int main() { while( ~scanf("%d %d", &n,…

题目:Click here #include <bits/stdc++.h> using namespace std; typedef long long ll; const int INF = 0x3f3f3f3f; +; int n, m; int main() { while( ~scanf("%d %d", &n, &m ) ) { ) { printf("1\n"); continue; } ) printf( ); else…

题目:Click here 题意:(据说这个题的题意坑了不少人啊~~~)题目一共给了3个数---- T 表示歌曲的长度(s).S 表示下载了歌曲的S后开始第一次播放(也就是说S秒的歌曲是事先下载好的).q 表示下载速度(每秒下载歌曲(q-1)/q秒).问题就是播放的速度比下载的速度快,每当播放到没下载的位置,就会重新从头播放,输出的就是从头播放的次数(包括第一次). 分析:高中物理追击问题,模拟下好了. #include <bits/stdc++.h> using namespace std;…

B. Simple Game time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to n. Let'…

C. Replacement time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Daniel has a string s, consisting of lowercase English letters and period signs (characters '.'). Let's define the operation…

B. Simple Game time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to n. Let'…

官方题解是离线询问,dfs树形转线性,然后二分找区间. 还有一种比较好的做法是直接dfs,将当前访问这个结点u相关的询问之前的状态存起来,然后访问完以后利用异或开关性,得到这颗子树上的答案. 代码是学习别人的http://blog.csdn.net/squee_spoon/article/details/47666667 #include<bits/stdc++.h> using namespace std; ; char s[maxn]; int head[maxn],to[maxn],nx…

先扫描一遍得到每个位置向后连续的'.'的长度,包含自身,然后在扫一遍求出初始的合并次数. 对于询问,只要对应位置判断一下是不是'.',以及周围的情况. #include<bits/stdc++.h> using namespace std; ; char s[maxn]; int post[maxn]; int main() { //freopen("in.txt","r",stdin); int n,m; scanf("%d%d",…

贪心,如果m分成的两个区间长度不相等,那么选长的那个区间最接近m的位置,否则选m-1位置,特判一下n等于1的情况 #include<bits/stdc++.h> using namespace std; int main() { int n,m; scanf("%d%d",&n,&m); ){ printf(; } , d2 = n-m; if(d1<d2){ printf(); }else { printf(); } ; }…

C. Replacement time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Daniel has a string s, consisting of lowercase English letters and period signs (characters '.'). Let's define the operation…

E. Pig and Palindromes   Peppa the Pig was walking and walked into the forest. What a strange coincidence! The forest has the shape of a rectangle, consisting of n rows and m columns. We enumerate the rows of the rectangle from top to bottom with num…

题目链接 题意:对于m次询问 求解以vi为根节点 深度为hi的的字母能不能组合成回文串. 思路:暴力dsu找一边 简直就是神技! #include<bits/stdc++.h> #define ll long long int using namespace std; inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;} int moth[13]={0,31…

Codeforces Round #748 (Div. 3) A. Elections 思路分析: 令当前值比最大值大即可,如果最大值是它自己,就输出\(0\) 代码 #include <bits/stdc++.h> using namespace std; pair<int, int> a[3]; int ans[3]; int main() { ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); int t; cin >&g…

Codeforces Round 363 Div. 1 题目链接:## 点击打开链接 A. Vacations (1s, 256MB) 题目大意:给定连续 \(n\) 天,每天为如下四种状态之一: 不能进行运动或比赛 可以进行运动但不能比赛 可以进行比赛但不能运动 可以进行比赛或运动 对于每天,可以根据当天的状态选择运动,比赛或休息.但不能连续两天的选择均为运动或均为比赛.求在这 \(n\) 天中最少需要休息多少天. 数据范围:\(n \leq 100\) 简要题解:令 \(f_{i,0},f_…

Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate it n = int(raw_input()) s = "" a = ["I hate that ","I love that ", "I hate it","I love it"] for i in ran…

Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…

直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输出”#Color”,如果只有”G”,”B”,”W”就输出”#Black&White”. #include <cstdio> #include <cstring> using namespace std; const int maxn = 200; const int INF =…