ACdream 1154 Lowbit Sum
2014-07-31 16:46
447 查看
Lowbit Sum
Time Limit: 2000/1000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others)SubmitStatisticNext
Problem
Problem Description
long long ans = 0;for(int i = 1; i <= n; i ++)
ans += lowbit(i)
lowbit(i)的意思是将i转化成二进制数之后,只保留最低位的1及其后面的0,截断前面的内容,然后再转成10进制数
比如lowbit(7),7的二进制位是111,lowbit(7) = 1
6 = 110(2),lowbit(6) = 2,同理lowbit(4) = 4,lowbit(12) = 4,lowbit(2) = 2,lowbit(8) = 8
每输入一个n,求ans
Input
多组数据,每组数据一个n(1 <= n <= 10^9)
Output
每组数据输出一行,对应的ans
Sample Input
1 2 3
Sample Output
1 3 4
Source
dream
Manager
scaucontest这道题,简直闹心啊,明明是一道水题,我花了好长时间才想出来---看来我还真是水的掉渣
下面是我的AC代码:
相关文章推荐
- ACdream 1154 Lowbit Sum (数位DP)
- acdream 1154 Lowbit Sum
- ACdream 1154 Lowbit Sum(数学:推理)
- ACdream 1154 Lowbit Sum (数位DP)
- ACdream 1154 Lowbit Sum (数位dp)
- ACdream 1154 Lowbit Sum(数位DP)
- ACdreamOJ 1154 Lowbit Sum (数字dp)
- 数位DP ACdream 1154 Lowbit Sum
- ACdream OJ 1154 Lowbit Sum
- ACDream - Lowbit Sum
- ACdreamOJ 1154 Lowbit Sum (数位dp)
- Lowbit Sum
- [LeetCode]338 Counting Bits(dp,lowbit)
- BZOJ 2064 浅谈状态压缩动态规划基础及lowbit枚举子集和
- acdream 1431 Sum vs Product
- ACDream - Divide Sum
- Codeforces Round #384(Div. 2)B. Chloe and the sequence【思维+lowbit】
- lowbit
- CF 792D - Paths in a Complete Binary Tree lowbit,模拟
- HDU 5975 Aninteresting game ( lowbit理解 )