Codeforces 768B Code For 1】的更多相关文章

768B - Code For 1 思路:类似于线段树的区间查询. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define mem(a,b) memset((a),(b),sizeof(a)) int query(ll L,ll R,ll n,ll l,ll r) { ; ); ll m=(l+r)>>; ,l,m-)+query(L,R,n%,m,m)+query(L,R,n/,m…
B. Code For 1 time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Jon fought bravely to rescue the wildlings who were attacked by the white-walkers at Hardhome. On his arrival, Sam tells him t…
Codeforces Avito Code Challenge 2018 D. Bookshelves 题目连接: http://codeforces.com/contest/981/problem/D Description Mr Keks is a typical white-collar in Byteland. He has a bookshelf in his office with some books on it, each book has an integer positive…
参考自:https://www.cnblogs.com/ECJTUACM-873284962/p/6423483.html B. Code For 1 time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Jon fought bravely to rescue the wildlings who were attacked by…
[题目链接]:http://codeforces.com/contest/768/problem/B [题意] 一开始给你一个数字n; 让你用这个数字n根据一定的规则生成序列; (如果新生成的序列里面还有大于1的数字,就一直按着上面的规则重复生成); 最后让你统计在一个区间范围内的1的数目; [题解] 一个树形的样子; 算出总共1的数目(整棵树的叶子节点上和节点的余数上) 这个挺好算的; 然后在从下往上走的时候记录每个节点的子树的size; 和子树所含的1的个数; 不 应该先算出总的size;…
这题给的一个教训:Codeforces没有超时这个概念.本来以为1000*(1000+1)/2*10*10要超时的.结果我想多了. 这题由于k层都可能有关系,所以建一个图,每两个点之间连边,边权为n*m和他们之间的差值*w的最小值,然后求一个最小生成树就可以得出结果.且可以证明不会存在环.由于边比较稠密,用Prim算法求最小生成树. 代码: #include <iostream> #include <cstdio> #include <cstring> #include…
Portal A. Antipalindrome 暴力. B. Businessmen Problems 暴力. C. Useful Decomposition 居然不是C打头的?! 将一棵树划分成若干条边不相交的路径,使得任意两个路径均有交点. 易知树上的两条路径最多有一个交点.若有三条路径两两相交,则必形成三个交点或一个交点.设路径1与路径2交于\(v_1\),路径1与路径3交于\(v_2\),路径2与路径3交于\(v_3\).若\(v_1\neq v_2 \neq v_3\),则有路径\(…
传送门 Description You are given names of two days of the week. Please, determine whether it is possible that during some non-leap year the first day of some month was equal to the first day of the week you are given, while the first day of the next mon…
传送门 Description You are given a table consisting of n rows and m columns. Numbers in each row form a permutation of integers from 1 to m. You are allowed to pick two elements in one row and swap them, but no more than once for each row. Also, no more…
拖了好久的题解,想想还是补一下吧. A. King of Thieves 直接枚举起点和5个点之间的间距,进行判断即可. #include <bits/stdc++.h> using namespace std; ]; int main() { //freopen("in.txt", "r", stdin); int n; bool ans = false; scanf("%d%s", &n, s); ; q < n &…