B. Little Artem and Dance time limit per test 2 second memory limit per test 256 megabytes input standard input output standard output Little Artem is fond of dancing. Most of all dances Artem likes rueda - Cuban dance that is danced by pairs of boys…

题目链接: Codeforces 669D Little Artem and Dance 题目描述: 给一个从1到n的连续序列,有两种操作: 1:序列整体向后移动x个位置, 2:序列中相邻的奇偶位置互换. 问:q次操作后,输出改变后的序列? 解题思路: 刚开始只看了第一组样例,发现相邻的奇偶位一直在一起,于是乎就开始writing code,写完后发现并不是正解!!!!就去推了一下第三个样例,总是这组实例通过,那组实例卡死,,,,,,,最后终于成功的Mengbility.今天突然想起来,其实整体…

http://codeforces.com/problemset/problem/669/D 题意:n个数1~N围成一个圈.q个操作包括操作1:输入x, 所有数右移x.操作2:1,2位置上的数(swap(a[1], a[2])交换:3,4交换.... 题解:观察,发现所有奇数行为都是一样的,偶数同理,(对于操作1 两者相同,对于操作2 奇数位++,偶数位上--: 模拟1,2,即可,然后依次输出其它数字即可.(看清 ac代码: #define _CRT_SECURE_NO_WARNINGS #in…

模拟. 每个奇数走的步长都是一样的,每个偶数走的步长也是一样的. 记$num1$表示奇数走的步数,$num2$表示偶数走的步数.每次操作更新一下$num1$,$num2$.最后输出. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #i…

D. Little Artem and Dance 题目连接: http://www.codeforces.com/contest/669/problem/D Description Little Artem is fond of dancing. Most of all dances Artem likes rueda - Cuban dance that is danced by pairs of boys and girls forming a circle and dancing tog…

题目链接: D. Little Artem and Dance time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little Artem is fond of dancing. Most of all dances Artem likes rueda — Cuban dance that is danced by pairs…

题目链接:http://codeforces.com/problemset/problem/669/D D. Little Artem and Dance time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little Artem is fond of dancing. Most of all dances Artem like…

D. Little Artem and Dance Little Artem is fond of dancing. Most of all dances Artem likes rueda - Cuban dance that is danced by pairs of boys and girls forming a circle and dancing together. More detailed, there are n pairs of boys and girls standing…

题目链接: http://codeforces.com/contest/669/problem/D 题意: 给你一个初始序列:1,2,3,...,n. 现在有两种操作: 1.循环左移,循环右移. 2.1,2位置交换,3,4位置交换,...,n-1,n位置交换 现在问执行了q次操作之后序列是什么,每次操作可以是两种操作的任意一种 题解: 我们把数列按位置的奇偶分为两堆,无论哪种操作,始终都还是这两堆,最多就是整堆的对换和一个堆内部的偏移. 所以我们只要记录第一个位置和第二个位置的数的变化(相当于每…

题意:2种操作,转动或者奇偶位互换. 不论怎么交换,1的后两位一定是3,3的后两位一定是5.因此只要记录1,2的位置. //#pragma comment(linker,"/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<string> #include<cstring> #include<vector> #include&l…