题目链接:http://codeforces.com/problemset/problem/669/D
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 in a circle. Initially, boy number 1 dances with a girl number 1, boy number 2 dances with a girl number 2 and so on. Girls are numbered in the clockwise order. During the dance different moves are announced and all pairs perform this moves. While performing moves boys move along the circle, while girls always stay at their initial position. For the purpose of this problem we consider two different types of moves:
- Value x and some direction are announced, and all boys move x positions in the corresponding direction.
- Boys dancing with even-indexed girls swap positions with boys who are dancing with odd-indexed girls. That is the one who was dancing with the girl 1 swaps with the one who was dancing with the girl number 2, while the one who was dancing with girl number 3 swaps with the one who was dancing with the girl number 4 and so one. It‘s guaranteed that n is even.
Your task is to determine the final position of each boy.
Input
The first line of the input contains two integers n and q (2?≤?n?≤?1?000?000, 1?≤?q?≤?2?000?000) — the number of couples in the rueda and the number of commands to perform, respectively. It‘s guaranteed that n is even.
Next q lines contain the descriptions of the commands. Each command has type as the integer 1 or 2 first. Command of the first type is given as x (?-?n?≤?x?≤?n), where 0?≤?x?≤?n means all boys moves x girls in clockwise direction, while ?-?x means all boys move x positions in counter-clockwise direction. There is no other input for commands of the second type.
Output
Output n integers, the i-th of them should be equal to the index of boy the i-th girl is dancing with after performing all q moves.
Examples
Input
6 31 221 2
Output
4 3 6 5 2 1
Input
2 31 121 -2
Output
1 2
Input
4 221 3
Output
1 4 3 2题目大意:n对男生女生顺时针围着一个圈圈在跳舞,一号男生和一号女生跳,依次类推。现在给几个指令,让男生改变他们的位置,而女生位置不变,要你求出执行完所有指令之后一号女生对应的是几号男生,二号女生对应几号男生,以此类推全部输出就行了。指令分为以下两种: 1.整体绕着这个圆圈走x步(正数为顺时针,负数为逆时针) 2.和位置数为奇数的女生跳舞的那个男生要和和位置数为偶数女生跳舞的那个男生交换以下位置,比如和一号女生跳舞的男生要和二号女生跳舞的那个男生交换位置,和3号女生跳舞的那个男生要和4号女生跳舞的那个男生交换位置,以此类推。 解题思路:这个题目看起来挺简单的,就是每次输入一个指令执行完相应的操作就行了的,可是呢,我们来看那一下范围:2?≤?n?≤?1?000?000, 1?≤?q?≤?2?000?000(n为人数,q为指令数),这么大的区间,如果我们一个一个执行操作的话,需要好多次循环,这样肯定是会超时的,所以我们想是否有什么巧妙的方法可以一次性知道他们最后的位置会怎样呢?可能先想到得应该是记录下执行了几次1操作执行了几次2操作,然后根据1操作的次数和2操作的次数来确定最后他们的位置吧,但是这样确实错的,因为先交换位置再整体移动和先整体移动在交换是不一样的,反正很乱,而且也不行,自己带几组数据试试就知道了。这时候,我们就仔细分析下两个操作的特点,第一个操作就是整体移动多少个位置,而第二个操作就是简单两个人交换位置,总人数为偶数,仔细想想我们就可以发现奇数位置的男生的相对位置压根就是不改变的,偶数位置的男生也是如此,意思就是1 3 5 7 9 ……这些男生的相对位置不会发生改变,同样2 4 6 8 ……也是如此。那题目就可以变得简单了,我们直接从开始几率1号男生和2号男生的位置就好了,就相当于记录了全部的奇数位置的男生和偶数位置的男生了,每次操作,我们对1号男生和2号男生进行更新即可,全部指令执行完毕后,我们通过1号2号男生的位置就可以填不上其他男生的位置了。思路就是这样子的,不过感觉好像不是很简单想出来那。。。。附上AC代码:
1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 int n,q,k,cnt; 5 int d[1000005]; 6 7 int main() 8 { 9 scanf("%d%d",&n,&q); 10 int fir=0,sec=1; //用来记录1号男生和2号男生的位置 11 while(q--) 12 { 13 scanf("%d",&k); 14 if(k==1) 15 { 16 scanf("%d",&cnt); //整体移动cnt个位置,注意是个圈 17 fir=(fir+cnt+n)%n; 18 sec=(sec+cnt+n)%n; 19 } 20 else 21 { 22 if(fir%2==0) //交换位置,先判断两个人的相对位置 23 { 24 fir=(fir+1+n)%n; 25 sec=(sec-1+n)%n; 26 } 27 else 28 { 29 fir=(fir-1+n)%n; 30 sec=(sec+1+n)%n; 31 } 32 } 33 } 34 for(int i=0;i<n;i+=2) 35 { 36 d[(fir+i)%n]=i+1; //奇数列男生位置填充 37 d[(sec+i)%n]=i+2; //偶数列男生位置填充 38 } 39 for(int i=0;i<n;i++) 40 { 41 if(i==0) printf("%d",d[i]); 42 else printf(" %d",d[i]); 43 } 44 printf("\n"); 45 return 0; 46 }
原文地址:https://www.cnblogs.com/zjl192628928/p/9350243.html