Codeforces Round #449 (Div. 2) https://codeforces.com/contest/897 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emplace_back…

http://codeforces.com/contest/896/problem/B 题意: 交互题 n张卡片填m个1到c之间的数,1<=n*ceil(c/2)<=m 最后填出一个单调非降序列,输出每次填的位置 <=c/2: 从左开始扫描,遇到空位 或 比原数更优(<原数)就放 >c/2: 从右开始扫描,遇到空位 或 比原数更优(>原数)就放 以<=c/2为例: 空位就放,<原数就覆盖,>=原数就往后扫 这样每个位置最多只会被覆盖c/2次,(c/2…

又掉分了0 0. A. Scarborough Fair time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Are you going to Scarborough Fair? Parsley, sage, rosemary and thyme. Remember me to one who lives there. He on…

http://codeforces.com/problemset/problem/896/C 题意: 对于一个随机序列,执行以下操作: 区间赋值 区间加 区间求第k小 区间求k次幂的和 对于随机序列,可以使用Old Driver Tree 就是将序列中,连续的相同值域合并为一段 然后暴力操作 #include<set> #include<vector> #include<cstdio> #include<iostream> #include<algor…

http://codeforces.com/contest/896/problem/A 第i个字符串嵌套第i-1个字符串 求第n个字符串的第k个字母 dfs #include<map> #include<cstdio> #include<iostream> using namespace std; typedef long long LL; ]={" What are you doing at the end of the world? Are you bus…

B. Chtholly's request time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output — Thanks a lot for today. — I experienced so many great things. — You gave me memories like dreams... But I have to le…

PROBLEM C/A. Nephren gives a riddle 题 http://codeforces.com/contest/896/problem/A codeforces 896a 897c 解 预第i个串的长度,由于k最大1e18,所以超过1e18的长度直接标记为无限大,不用计算 然后递归搜索即可 #include <iostream> #include <cstring> #include <cstdio> #include <cmath>…

A. Scarborough Fair time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Are you going to Scarborough Fair? Parsley, sage, rosemary and thyme. Remember me to one who lives there. He once was th…

题目链接 交互题. 题意:给你三个数n,m,k.让你完成至多m次互动,每次给你一个q,让你从n个位置选一个位置放这个数,覆盖已经放过的数.让你再m次使得n个位置的数不递减,达到直接退出. 解法:暴力,如果q小于c/2的话,从前往后找,若当前位置没有数或者比q大的话,就直接放再这个位置上,大于c/2的话就从后往前找,若当前位置没有数或者比q小的话就直接放. 如果每个位置都有数的话说明放完了. #include<bits/stdc++.h> #define LL long long #define…

B:注意到nc/2<=m,于是以c/2为界决定数放在左边还是右边,保证序列满足性质的前提下替换掉一个数使得其更靠近边界即可. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #de…