链接:https://codeforces.com/contest/1100/problem/A 题意: 给定n,k. 给定一串由正负1组成的数. 任选b,c = b + i*k(i为任意整数).将c所有c位置的数删除,求-1和1个数差值绝对值的最大值. 思路: 暴力遍历 代码: #include <bits/stdc++.h> using namespace std; int a[110]; int main() { int n,k; scanf("%d%d",&…

Codeforces Round #532 (Div. 2) A - Roman and Browser #include<bits/stdc++.h> #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> #include<queue> #inclu…

Codeforces Round #532 (Div. 2) 题目总链接:https://codeforces.com/contest/1100 A. Roman and Browser 题意: 给出由-1和1组成的n个数,现在任意选定一个起点,从起点开始向左向右k个k个地拿走.最后问abs(cnt(-1)-cnt(1))最大是多少. 题解: 由于n和k最多只有100,所以枚举起点直接暴力就好了. 代码如下: #include <bits/stdc++.h> using namespace s…

A. Roman and Browser 签到. #include <bits/stdc++.h> using namespace std; ]; int get(int b) { ]; memset(vis, , sizeof vis); int c = b; while (c <= n) { vis[c] = ; c += k; } c = b - k; ) { vis[c] = ; c -= k; } , r = ; ; i <= n; ++i) if (!vis[i]) {…

Roman and Numbers time limit per test 4 seconds memory limit per test 512 megabytes input standard input output standard output Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change his…

题目链接: http://codeforces.com/problemset/problem/401/D D. Roman and Numbers time limit per test4 secondsmemory limit per test512 megabytes 问题描述 Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sere…

D. Roman and Numbers time limit per test 4 seconds memory limit per test 512 megabytes input standard input output standard output Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change h…

https://codeforces.com/contest/1100/problem/F 题意 一个有n个数组c[],q次询问,每次询问一个区间的子集最大异或和 题解 单问区间子集最大异或和,线性基能处理,但是这次多次询问,假如每次重新建立基向量会超时 考虑区间的优先级,假如我只插入不删除的话,区间的优先级和左端点没有关系 贪心一下,只保留后面插入的基,这样就可以离线解决询问,然后查询的时候需要判基的位置是不是在左端点后面 代码 #include<bits/stdc++.h> #define…

F. Ivan and Burgers 题目链接:https://codeforces.com/contest/1100/problem/F 题意: 给出n个数,然后有多个询问,每次回答询问所给出的区间的异或和最大值. 题解: 考虑离线做法,先把所有的询问区间按照右端点进行排序,然后从1开始逐个将ai插入,插入线性基的同时记录一下位置,最后扫一下,看看哪些的位置是不小于li的即可加入答案. 这种做法在时间复杂度上面是可行的,但是需要注意的是,如果在i这个位置插入最高位为x的线性基,同时在j这个位…

D. Roman Digits time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers…