Lowbit Sum

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

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

Information
Solved Number564
Submit Number1318
Problem Tags
dp
math
No tag edit access
温馨提示:AC后可以编辑标签哦. ^-^
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