传送门:点我 You have an array a consisting of n integers. Each integer from 1 to n appears exactly once in this array. For some indices i (1 ≤ i ≤ n - 1) it is possible to swap i-th element with (i + 1)-th, for other indices it is not possible. You may pe…

Educational Codeforces Round 37 (Rated for Div. 2)C. Swap Adjacent Elements time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have an array a consisting of n integers. Each integer from 1…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 显然l..r这一段连续的1可以把l..r+1变成有序的. 那么就把所有的连续1段变成有序的就好. 看看最后是不是升序即可. [代码] #include <bits/stdc++.h> using namespace std; const int N = 2e5; int n,a[N+10],b[N+10]; char s[N+10]; int main(){ #ifdef LOCAL_DEFINE freopen("…

You have an array a consisting of n integers. Each integer from 1 to n appears exactly once in this array. For some indices i (1 ≤ i ≤ n - 1) it is possible to swap i-th element with (i + 1)-th, for other indices it is not possible. You may perform a…

我也不知道该说啥,水就是了~ code: #include <bits/stdc++.h> #define N 300004 #define setIO(s) freopen(s".in","r",stdin) using namespace std; char S[N]; int a[N],s[N]; int main() { // setIO("input"); int i,j,n,flag=0; scanf("%d&q…

原题链接:http://codeforces.com/gym/100203/attachments/download/1702/statements.pdf 题解 考虑暴力的复杂度是O(n^3),所以我们需要记录所有的ai+aj,如果当前考虑到了ak,那么就去前面寻找ai,使得ak-ai是我们记录过的和.整个算法的复杂度O(n^2). 代码 #include<iostream> #include<cstring> #include<cstdio> #include<…

In a string composed of 'L', 'R', and 'X'characters, like "RXXLRXRXL", a move consists of either replacing one occurrence of "XL" with "LX", or replacing one occurrence of "RX" with "XR". Given the startin…

In a string composed of 'L', 'R', and 'X' characters, like "RXXLRXRXL", a move consists of either replacing one occurrence of "XL" with "LX", or replacing one occurrence of "RX" with "XR". Given the starti…

题意 给你一个1-n的排列. 并给你一个字符串——其中用0和1表示对应数列中的位置上的值可不可以和后面相邻的数交换. 判断该数列能否在限制中交换为不降序数列. 思路 由于刚学了树状数组,一开始以为是用这样的数据结构去找有没有逆序. 事实上题目中的1~n并且每个数并不相同应该引起注意.关键就是这了. 下面是两种不同的思路: #include <cstdio> #include <iostream> #include <cstring> #include <strin…

传送门 好像是个挺显然的贪心 首先每次交换当然要尽量一次交换就多两个相同的位置 即 优先把 $\begin{bmatrix}a\\ b\end{bmatrix}$ 和 $\begin{bmatrix}a\\ b\end{bmatrix}$ 交换 优先把 $\begin{bmatrix}b\\ a\end{bmatrix}$ 和 $\begin{bmatrix}b\\ a\end{bmatrix}$ 交换 最后如果剩下$\begin{bmatrix}a\\ b\end{bmatrix}$,$\be…