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…

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…

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 <…

Beru-taxi 题目链接: http://codeforces.com/contest/706/problem/A Description Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi…

Codeforces Round #603 (Div. 2) A. Sweet Problem A. Sweet Problem time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have three piles of candies: red, green and blue candies: the first pile…

A. Uncowed Forces Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/604/problem/A Description Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maxi…

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…