Problem   Codeforces Round #539 (Div. 2) - D. Sasha and One More Name Time Limit: 1000 mSec Problem Description Input The first line contains one string s (1≤|s|≤5000) — the initial name, which consists only of lowercase Latin letters. It is guarante…

Problem   Codeforces Round #539 (Div. 2) - C. Sasha and a Bit of Relax Time Limit: 2000 mSec Problem Description Input The first line contains one integer n (2≤n≤3⋅10^5) — the size of the array. The second line contains n integers a1,a2,…,an (0≤ai<2^…

Codeforces Round #539 (Div. 2) A - Sasha and His Trip #include<bits/stdc++.h> #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> #include<queue> #incl…

Codeforces Round #539 (Div. 2) 题目链接:https://codeforces.com/contest/1113 A. Sasha and His Trip 题意: n个城市,城市分布在一条直线上且按升序排序,现在有个人开车从一号城市出发,车的油箱容量为v. 在每个城市都可以买油,但价格不一样:第i个城市买1单位的油花费i元.问最终从1到n花费的最少为多少. 题解: 贪心即可,尽量在前面的城市买油,最后一鼓作气到n号城市. 代码如下: #include <bits/…

https://codeforces.com/contest/1113/problem/D 题意 将一个回文串切成一段一段,重新拼接,组成一个新的回文串,问最少切几刀 题解 首先无论奇偶串,最多只会切两刀 然后对于偶数串,看看有没有循环回文串,有的话只需要切一刀 代码 #include<bits/stdc++.h> using namespace std; int n,m,i,j,ok,ans; string s; int main(){ cin>>s;n=s.size(); if…

https://codeforces.com/contest/1113/problem/C 题意 一个n个数字的数组a[],求有多少对l,r满足\(sum[l,mid]=sum[mid+1,r]\),sum为异或和(n<=3e5,a[i]<=2^20) 题解 异或和为零的区间可以分成任意两个区间(这两个区间的异或和相等) 定义dp[i][j]为异或和为i,下标为j(只记录奇偶)的前缀个数 枚举r,然后累加l 代码 #include<bits/stdc++.h> #define M…

转载自:https://blog.csdn.net/Charles_Zaqdt/article/details/87522917 题目链接:https://codeforces.com/contest/1113/problem/C        题意是给了n个数字,让找出一个长度为偶数的区间[l, r],使得al ^ al+1 ^ .... ^ amid = amid + 1 ^ ... ^ ar这个等式成立(l到mid的异或和等于mid+1到r的异或和),求出有多少个满足要求的区间.    …

题中意思显而易见,即求满足al⊕al+1⊕…⊕amid=amid+1⊕amid+2⊕…⊕ar且l到r的区间长为偶数的这样的数对(l,r)的个数. 若al⊕al+1⊕…⊕amid=amid+1⊕amid+2⊕…⊕ar,我们可以推出al⊕al+1⊕…⊕amiamid+1⊕amid+2⊕…⊕ar=0:反推也是可以成立的. 我们已知任何数0对异或都等于本身.所以当前数异或一段数之后等于本身,那么异或之后的这段数肯定是异或为0的,我们只需要知道这一段是不是长度为偶数即可. 我们从头异或一道,若异或到某个数…

A:即求长度为偶数的异或和为0的区间个数,对前缀异或和用桶记录即可. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 300010 char getc(){char…

这场怎么全是数据结构题...…