C. Find Amir time limit per test   1 second memory limit per test   256 megabytes   A few years ago Sajjad left his school and register to another one due to security reasons. Now he wishes to find Amir, one of his schoolmates and good friends. There…

1 2 3 4 5 6 7 4-5-3-6-2-7-1 答案是(n-1)/2 #include<cstdio> using namespace std; int n; int main(){ scanf("%d",&n); printf("%d\n",(n-1)/2); return 0; }…

A. Fake NP time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path.…

来自FallDream的博客,未经允许,请勿转载,谢谢. 由于人傻又菜 所以这次又滚去div2了  一堆结论题真的可怕 看见E题不是很有思路   然后就去大力搞F题  T了最后一个点 真的绝望   但是还是上紫了 感觉每次cf都在乱打 我好菜啊都不会 A.Fake NP 给定l和r,求[l,r]中的数的因数中出现次数最多的那一个 sb题 如果l和r相同输出它 不然输出2 #include<iostream> #include<cstdio> #include<cstring&…

A. Fake NP 题意:给你l,r,让你输出[l,r]里面除1以外的,出现因子数量最多的那个数. 题解:如果l==r输出l,否则都输出2 #include<bits/stdc++.h> using namespace std; int main(){ int l,r; cin>>l>>r; if(r-l==0){ cout<<l<<endl; return 0; } int ans = 0; int num = 0; for(int i=2;…

比赛时候切了A-E,fst了A Standings第一页只有三个人挂了A题,而我就是其中之一,真™开心啊蛤蛤蛤 A. Fake NP time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak…

A Fake NP standard input/output s, MB Submit Add to favourites x3673 B -palindrome standard input/output s, MB Submit Add to favourites x3760 C Find Amir standard input/output s, MB Submit Add to favourites x3503 D Minimum number of steps standard in…

题目大意:给出一个森林,每次询问给出u,v,问从u所在连通块中随机选出一个点与v所在连通块中随机选出一个点相连,连出的树的直径期望(不是树输出-1).(n,q<=10^5) 解法:预处理出各连通块的直径和各点到连通块内一点的最远距离d[x](树形dp+换根),询问若在同一块内输出-1,否则若随机选出两点x,y,直径为max(d[x]+d[y]+1,x所在块直径,y所在块直径),我们把同一连通块内的d排序,枚举小的连通块中的d,到大的连通块中二分d[x]+d[y]+1<=max(x所在块直径,y…

D. Minimum number of steps time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of s…

对那个树进行dfs,在动态维护那个当前的冰激凌集合的时候,显然某种冰激凌仅会进出集合各一次(因为在树上形成连通块). 于是显然可以对当前的冰激凌集合贪心染色.暴力去维护即可.具体实现看代码.map不必要. #include<cstdio> #include<set> #include<vector> #include<map> #include<algorithm> using namespace std; map<int,bool>…