hdu 4349
2015-08-03 15:35
162 查看
Xiao Ming's Hope
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1627 Accepted Submission(s): 1091
[align=left]Problem Description[/align]
Xiao Ming likes counting numbers very much, especially he is fond of counting odd numbers. Maybe he thinks it is the best way to show he is alone without a girl friend. The day 2011.11.11 comes. Seeing classmates walking with their girl friends, he coundn't help running into his classroom, and then opened his maths book preparing to count odd numbers. He looked at his book, then he found a question "C(n,0)+C(n,1)+C(n,2)+...+C(n,n)=?". Of course, Xiao Ming knew the answer, but he didn't care about that , What he wanted to know was that how many odd numbers there were? Then he began to count odd numbers. When n is equal to 1, C(1,0)=C(1,1)=1, there are 2 odd numbers. When n is equal to 2, C(2,0)=C(2,2)=1, there are 2 odd numbers...... Suddenly, he found a girl was watching him counting odd numbers. In order to show his gifts on maths, he wrote several big numbers what n would be equal to, but he found it was impossible to finished his tasks, then he sent a piece of information to you, and wanted you a excellent programmer to help him, he really didn't want to let her down. Can you help him?
[align=left]Input[/align]
Each line contains a integer n(1<=n<=108)
[align=left]Output[/align]
A single line with the number of odd numbers of C(n,0),C(n,1),C(n,2)...C(n,n).
[align=left]Sample Input[/align]
1
2
11
[align=left]Sample Output[/align]
2
2
8
[align=left]Author[/align]
HIT
[align=left]Source[/align]
2012 Multi-University Training Contest 5
[align=left]Recommend[/align]
zhuyuanchen520
lucas定理,并没有听过,但是感觉好厉害的样子....
/************************************************************************* > File Name: code/whust/#8/E.cpp > Author: 111qqz > Email: rkz2013@126.com > Created Time: 2015年08月03日 星期一 15时26分58秒 ************************************************************************/ #include<iostream> #include<iomanip> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> #include<map> #include<set> #include<queue> #include<vector> #include<stack> #define y0 abc111qqz #define y1 hust111qqz #define yn hez111qqz #define j1 cute111qqz #define tm crazy111qqz #define lr dying111qqz using namespace std; #define REP(i, n) for (int i=0;i<int(n);++i) typedef long long LL; typedef unsigned long long ULL; const int inf = 0x7fffffff; int main() { int n; while (~scanf("%d",&n)) //Lucas定理,orzorz { int g = 1 ; int s = 0 ; while (n) { s = s + (n&1); n = n / 2; } for ( int i = 1 ; i <= s; i ++) { g = g * 2; } printf("%d\n",g); } return 0; }
相关文章推荐
- 事件的标准模式
- 百度地图api如何使用
- ubuntu学习日记
- 06-2. 字符串字母大小写转换(10)
- 【Linux】【Shell】Ubuntu Shell 前后台任务控制
- 【Linux】【Shell】Ubuntu Shell 前后台任务控制
- 【Linux】【Shell】Ubuntu Shell 前后台任务控制
- 【Linux】【Shell】Ubuntu Shell 前后台任务控制
- 【Linux】【Shell】Ubuntu Shell 前后台任务控制
- 【Linux】【Shell】Ubuntu Shell 前后台任务控制
- 【Linux】【Shell】Ubuntu Shell 前后台任务控制
- Android屏幕适配,百分比布局
- Ubuntu系统下eclipse配置mapreduce插件常见错误和解决办法汇总
- JavaScript Boolean(逻辑)对象
- 系统设计---笔记
- 10个CSS简写/优化技巧
- Apache -- 压力测试工具ab.exe
- access restriction
- JS触发服务器控件的单击事件
- 【转】对Java Serializable(序列化)的理解和总结