思路:很裸的康拓展开.. 我的平衡树居然跑的比树状数组+二分还慢.. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PII pair<int, int> #define y1 skldjfskldjg #define y2 skldfjsklejg using namespace std; ; ; const i…

题目传送门 /* 题意:给出无向无环图,每一个点的度数和相邻点的异或和(a^b^c^....) 图论/位运算:其实这题很简单.类似拓扑排序,先把度数为1的先入对,每一次少一个度数 关键在于更新异或和,精髓:a ^ b = c -> a ^ c = b, b ^ c = a; */ #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <…

题目传送门 /* 题意:给出一系列名字变化,问最后初始的名字变成了什么 字符串处理:每一次输入到之前的找相印的名字,若没有,则是初始的,pos[m] 数组记录初始位置 在每一次更新时都把初始pos加上去,那么就保证更新了初始的名字,这也是唯一要思考的地方了:) */ #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <cmath&…

题目传送门 /* 题意:给出无向无环图,每一个点的度数和相邻点的异或和(a^b^c^....) 图论/位运算:其实这题很简单.类似拓扑排序,先把度数为1的先入对,每一次少一个度数 关键在于更新异或和,精髓:a ^ b = c -> a ^ c = b, b ^ c = a; */ #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <…

题目链接: 题目 A. Misha and Forest time limit per test 1 second memory limit per test 256 megabytes 问题描述 Let's define a forest as a non-directed acyclic graph (also without loops and parallel edges). One day Misha played with the forest consisting of n ver…

传送门 Description Let's define a forest as a non-directed acyclic graph (also without loops and parallel edges). One day Misha played with the forest consisting of n vertices. For each vertex v from 0 to n - 1 he wrote down two integers, degreev and sv…

Codeforces Round #485 (Div. 2) E. Petr and Permutations 题目连接: http://codeforces.com/contest/987/problem/E Description Petr likes to come up with problems about randomly generated data. This time problem is about random permutation. He decided to gene…

A. Contest time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully sub…

A. Contest time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully sub…

题意:给出两个排列,求出每个排列在全排列的排行,相加,模上n!(全排列个数)得出一个数k,求出排行为k的排列. 解法:首先要得出定位方法,即知道某个排列是第几个排列.比如 (0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 2, 0), (2, 0, 1), (2, 1, 0). 拿排列(1,2,0)来说,首位是1,前面有cnt=1个小于1的没被用过的数(0),所以它的排行要加上(cnt=1)*2!,第二位为2,因为1已经放了,所以小于2的只有0了,即cnt=1个,所以,排…