Codeforces Round #238 (Div. 2) D. Toy Sum(想法题)
Description
Little Chris is very keen on his toy blocks. His teacher, however, wants Chris to solve more problems, so he decided to play a trick on Chris.
There are exactly s blocks in Chris's set, each block has a unique number from 1 to s. Chris's teacher picks a subset of blocks X and keeps it to himself. He will give them back only if Chris can pick such a non-empty subset Y from the remaining blocks, that the equality holds:
"Are you kidding me?", asks Chris.
For example, consider a case where s = 8 and Chris's teacher took the blocks with numbers 1, 4 and 5. One way for Chris to choose a set is to pick the blocks with numbers 3 and 6, see figure. Then the required sums would be equal: (1 - 1) + (4 - 1) + (5 - 1) = (8 - 3) + (8 - 6) = 7.
However, now Chris has exactly s = 106 blocks. Given the set X of blocks his teacher chooses, help Chris to find the required set Y!
Input
The first line of input contains a single integer n (1 ≤ n ≤ 5·105), the number of blocks in the set X. The next line contains n distinct space-separated integers x1, x2, ..., xn (1 ≤ xi ≤ 106), the numbers of the blocks in X.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
In the first line of output print a single integer m (1 ≤ m ≤ 106 - n), the number of blocks in the set Y. In the next line output m distinct space-separated integers y1, y2, ..., ym (1 ≤ yi ≤ 106), such that the required equality holds. The sets X and Y should not intersect, i.e. xi ≠ yj for all i, j (1 ≤ i ≤ n; 1 ≤ j ≤ m). It is guaranteed that at least one solution always exists. If there are multiple solutions, output any of them.
Sample Input
31 4 5
11
Sample Output
2999993 1000000
11000000
思路
题意:
从 1 ~ 1000000 中选择 n 个数:x1,x2,...,xn,对 x1-1,x2-1,...xn-1 求和得s1。然后在 1 ~ 1000000 中除已经选择过的n个数中选择一些数,假设为y1, y2,...ym,设s = 1000000,对s-y1,s-y2,...,s-ym求和,如果这个和与s1相等,则输出y1,y2,...,ym
题解:
换个角度思考,由于集合X中:x1,x2,...,xn 是各不相同的,那么在S - X,设为Y(假定S是全集:1,2,...,n)对每个数xi(i : 1 ~ n)一定有相应的s-i+1与之对应(前提是,如果S-xi不在集合X中);如果有相应的s-xi+1在X中,那么可以找没有选择过的yj,s-yj+1来替换xi, s-xi+1。例如X中有 100, 999901而没有99, 999902,那么可以选择99, 999902来替代。效果是相同的。这样Y中的数量跟n是相同的。
官方题解:
Let's define the symmetric number of k to be s + 1 - k. Since in this case s is an even number, k ≠ s - k.
Note that (k - 1) + (s + 1 - k) = s, i.e., the sum of a number and its symmetric is always s. Let's process the given members x of X. There can be two cases:
- If the symmetric of x does not belong to X, we add it to Y. Both give equal values to the respective sums: x - 1 = s - (s + 1 - x).
- The symmetric of x belongs to X. Then we pick any y that neither y and symmetric of y belong to X, and add them to Y. Both pairs give equal values to the respective sums, namely s.
How to prove that in the second step we can always find such y? Let the number of symmetric pairs that were processed in the step 1 be a, then there remain other pairs. Among them, for pairs both members belong to X, and for other pairs none of the members belong to X. To be able to pick the same number of pairs for Y, as there are in X, we should have
which is equivalent to , as given in the statement.
Solution complexity: O(s) / O(n).
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int maxn = 1000000; int vis[1000005]; int main() { int N; while (~scanf("%d",&N)) { memset(vis,0,sizeof(vis)); int tmp,cnt = 0; bool first = true; for (int i = 0;i < N;i++) { scanf("%d",&tmp); vis[tmp] = 1; } printf("%d\n",N); for (int i = 1;i <= maxn;i++) { if (vis[i] && !vis[maxn+1-i]) { first?printf("%d",maxn+1-i):printf(" %d",maxn+1-i); first = false; cnt++; } } for (int i = 1;i <= maxn && cnt != N;i++) { if (!vis[i] && !vis[maxn+1-i]) { printf(" %d %d",i,maxn+1-i); cnt += 2; } } printf("\n"); } return 0; }
Codeforces Round #238 (Div. 2) D. Toy Sum(想法题)的更多相关文章
- Codeforces Round #238 (Div. 2) D. Toy Sum 暴搜
题目链接: 题目 D. Toy Sum time limit per test:1 second memory limit per test:256 megabytes 问题描述 Little Chr ...
- Codeforces Round #238 (Div. 2) D. Toy Sum
D. Toy Sum time limit per test:1 second memory limit per test:256 megabytes input:standard input o ...
- Codeforces Round #303 (Div. 2) A. Toy Cars 水题
A. Toy Cars Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/545/problem ...
- 水题 Codeforces Round #303 (Div. 2) A. Toy Cars
题目传送门 /* 题意:5种情况对应对应第i或j辆车翻了没 水题:其实就看对角线的上半边就可以了,vis判断,可惜WA了一次 3: if both cars turned over during th ...
- Codeforces Round #556 (Div. 2) - C. Prefix Sum Primes(思维)
Problem Codeforces Round #556 (Div. 2) - D. Three Religions Time Limit: 1000 mSec Problem Descripti ...
- Codeforces Round #575 (Div. 3) 昨天的div3 补题
Codeforces Round #575 (Div. 3) 这个div3打的太差了,心态都崩了. B. Odd Sum Segments B 题我就想了很久,这个题目我是找的奇数的个数,因为奇数想分 ...
- Codeforces Round #367 (Div. 2) A. Beru-taxi (水题)
Beru-taxi 题目链接: http://codeforces.com/contest/706/problem/A Description Vasiliy lives at point (a, b ...
- Codeforces Round #238 (Div. 1)
感觉这场题目有种似曾相识感觉,C题还没看,日后补上.一定要坚持做下去. A Unusual Product 题意: 给定一个n*n的01矩阵,3种操作, 1 i 将第i行翻转 2 i 将第i列翻转 3 ...
- Codeforces Round #334 (Div. 2) A. Uncowed Forces 水题
A. Uncowed Forces Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/604/pro ...
随机推荐
- 不能在DropDownList 中选择多个项
在绑定DropDownList时如果出现多次绑定,会出错以下错误: “不能在DropDownList 中选择多个项” 经了解,只需要在选中值是清空选择即可:xxDropDownList.ClearSe ...
- Ajax.ActionLink参数详解
该语法会生成一个a标签,点击a标签会执行一个Ajax请求. 有12个方法重载,下面详解方法中的各项参数: 参数一:linkText string类型 说明:链接显示的文字内容 参数二:actionNa ...
- T-SQL 将存储过程结果插入到表中
解决方案1: CREATE TABLE #tmpBus ( COL1 INT, COL2 INT ) INSERT INTO #tmpBus Exec SpGetRecords 'Pa ...
- 使用JNI封装底层input系统提供的event事件
首先说下思路,本文采用jni技术封装底层触摸事件,封装成MotionEvent类一样,不过没有android系统MotionEvent强大.源码MotionEvent位置:java-->fram ...
- Canvas绘图基础(一)
简单图形绘制 矩形:描边与填充 Canvas的API提供了三个方法,分别用于矩形的清除.描边及填充 clearRect(double x, double y, double w, double h) ...
- 固态硬盘与机械硬盘 SQL Server 单表插入性能对比测试
测试环境
- Makefile 编写 tips
1.变量赋值 VARIABLE = value #在执行时扩展,允许递归扩展 VARIABLE := value #在定义时扩展 VARIABLE ?= value #只有在该变量为空时才设置该值 V ...
- Linux下oracle环境变量无效问题
今天在维护oracle数据库时,查看监听的状态,执行 #lsnrctl status 报错: -bash:lsnrctl:command not found.以前并不会这样,仔细想了一下,问题找到了, ...
- vuejs mvvm图解
- ComponetArt Upload上传组件
componentArt的组件很炫,但示例并非单独的,要分离开来用还得费一番周折. 一个不经意问题,足足困扰了一好几天,服务器的时间不对,导致控件无法正常显示,无浏览文件按钮 看来,控件的开发者设定了 ...