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.    …

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^…

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…

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…

题解看这里 liouzhou_101的博客园 更简洁的代码看这里: #include <bits/stdc++.h> using namespace std; typedef long long LL; #define X first #define Y second inline void read(int &x) { int flag = 1; char ch; while(!isdigit(ch=getchar()))if(ch=='-')flag=-flag; for(x=0;…

如果mod是质数就好做了,但是做除法的时候对于合数mod可能没有逆元.所以就只有存一下mod的每个质因数(最多9个)的幂,和剩下一坨与mod互质的一部分.然后就能做了.有点恶心. CODE #include <bits/stdc++.h> using namespace std; typedef long long LL; const int MAXN = 100005; const int MAXP = 9; int n, q, a[MAXN], p[MAXP], cnt, mod, phi…

题解请看 Felix-Lee的CSDN博客 写的很好,不过最后不用判断最小值是不是1,因为[i,i]只有一个点,一定满足条件,最小值一定是1. CODE 写完就A,刺激. #include <bits/stdc++.h> using namespace std; typedef long long LL; #define X first #define Y second inline void read(int &x) { int flag = 1; char ch; while(!i…

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/…

E. Sasha and Array 题目连接: http://codeforces.com/contest/719/problem/E Description Sasha has an array of integers a1, a2, ..., an. You have to perform m queries. There might be queries of two types: 1 l r x - increase all integers on the segment from l…