Codeforces Round #250 (Div. 2)——The Child and Set
- 题意:
给定goal和limit,求1-limit中的若干个数,每一个数最多出现一次,且这些数的lowbit()值之和等于goal,假设存在这种一些数,输出个数和每一个数;否则-1 - 分析:
先考虑一下比較普通的情况,给一些数,和一个goal,问时候能达到。(最好还是设这些数已经从大到小排序)
考虑能否够贪心,对于当前的数x:
1、之后的数的和能等于x,那么假设x<=goal,显然必须选x;
2、之后的数的和能等于x-1,那么同上(这个情况就是二进制的情况)
3、之后的数的和不包含上述两个情况,那么不能贪心(推測)
再分析这个题,是符合第一个情况的。对于第i + 1位,当第i位出现两个1时候,之后才会在i + 1位出现一个1,所以符合第一个情况,能够贪心
lowbit()函数事实上就是一个数的二进制最低位的1代表的十进制数值
const int MAXN = 25; int lowbit(int n)
{
return n & -n;
} int bin(int n)
{
int ret = 0;
while (n)
{
n >>= 1;
ret++;
}
return ret - 1;
}
vector<int> G[MAXN]; int main()
{
// freopen("in.txt", "r", stdin);
int sum, limit;
while (~RII(sum, limit))
{
REP(i, MAXN) G[i].clear();
vector<int> ans;
int s = 0;
FE(i, 1, limit)
{
int t = bin(lowbit(i));
G[t].push_back(i);
s += lowbit(i);
}
if (s < sum)
puts("-1");
else
{
FED(i, 22, 0)
{
int val = (1 << i);
int ct = min((int)G[i].size(), sum / val);
REP(j, ct)
ans.push_back(G[i][j]);
sum -= ct * val;
}
WI(ans.size());
REP(i, ans.size())
cout << ans[i] << ' ';
puts("");
}
}
return 0;
}
Codeforces Round #250 (Div. 2)——The Child and Set的更多相关文章
- Codeforces Round #250 (Div. 1) D. The Child and Sequence 线段树 区间取摸
D. The Child and Sequence Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest ...
- Codeforces Round #250 (Div. 1) B. The Child and Zoo 并查集
B. The Child and Zoo Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/438/ ...
- Codeforces Round #250 (Div. 1) A. The Child and Toy 水题
A. The Child and Toy Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/438/ ...
- Codeforces Round #250 (Div. 1) D. The Child and Sequence (线段树)
题目链接:http://codeforces.com/problemset/problem/438/D 给你n个数,m个操作,1操作是查询l到r之间的和,2操作是将l到r之间大于等于x的数xor于x, ...
- Codeforces Round #250 (Div. 2)—A. The Child and Homework
好题啊,被HACK了.曾经做题都是人数越来越多.这次比赛 PASS人数 从2000直掉 1000人 被HACK 1000多人! ! ! ! 没见过的科技啊 1 2 4 8 这组数 被黑的 ...
- Codeforces Round #250 (Div. 1) D. The Child and Sequence(线段树)
D. The Child and Sequence time limit per test 4 seconds memory limit per test 256 megabytes input st ...
- Codeforces Round #250 (Div. 1) D. The Child and Sequence
D. The Child and Sequence time limit per test 4 seconds memory limit per test 256 megabytes input st ...
- Codeforces Round #250 (Div. 2) D. The Child and Zoo 并查集
D. The Child and Zoo time limit per test 2 seconds memory limit per test 256 megabytes input standar ...
- Codeforces Round #250 (Div. 2)B. The Child and Set 暴力
B. The Child and Set At the children's day, the child came to Picks's house, and messed his house ...
随机推荐
- mysql error: Access denied for user 'root'@'localhost' (using password: YES)
昨天重装了下系统,安装好mysql后,安装了客户端工具连接mysql,提示Access denied for user 'root'@'localhost' (using password: YES) ...
- koa 笔记 运行错误
按照 演示的代码 直接运行会出错,大家需要调整方式. http://koajs.cn/ 要安装以下 $ npm install -g n$ n 0.11.12$ node --harmony my-k ...
- Northwind数据库表字段介绍
① Categories:种类表 相应字段: CategoryID :类型ID: CategoryName:类型名; Description:类型说明; Picture:产品样本 ② Customer ...
- hive权威安装出现的不解错误!(完美解决)两种方法都可以
以下两种方法都可以,推荐用方法一! 方法一: 步骤一: yum -y install mysql-server 步骤二:service mysqld start 步骤三:mysql -u root - ...
- 【多线程】Java并发编程:Lock(转载)
原文链接:http://www.cnblogs.com/dolphin0520/p/3923167.html Java并发编程:Lock 在上一篇文章中我们讲到了如何使用关键字synchronized ...
- 【多线程】JAVA多线程和并发基础面试问答(转载)
JAVA多线程和并发基础面试问答 原文链接:http://ifeve.com/java-multi-threading-concurrency-interview-questions-with-ans ...
- Codeforces Round #367 (Div. 2) B. Interesting drink (模拟)
Interesting drink 题目链接: http://codeforces.com/contest/706/problem/B Description Vasiliy likes to res ...
- labview视频采集IMAdx
grab(连续采集) 摄像头打开之后便一直采集图像,存储在IMAQ开辟的临时空间里,只要while循环不断的读取临时空间就可以显示当前图像(grab调用的是image句柄)
- ASP.NET中Request.ApplicationPath、Request.FilePath、Request.Path、.Request.MapPath、Server.MapPath(转载)
1.Request.ApplicationPath->当前应用的目录 Jsp中, ApplicationPath指的是当前的application(应用程序)的目录,ASP.NET中也是这个 ...
- Top 7 Myths about HTTPS
Myth #7 – HTTPS Never Caches People often claim that HTTPS content is never cached by the browser; p ...