Lowbit Sum

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

123

Sample Output

134

/** this code is made by QuQ* Problem: 1154* Verdict: Accepted* Submission Date: 2014-07-31 11:59:02* Time: 188MS* Memory: 1676KB*/#include <iostream>#include <cmath> using namespace std; typedef long long LL;LL n, ans; int main(){    while (cin >> n) {        ans = 0;         if (n % 2 == 0)            ans += n / 2;        else            ans += (n / 2 + 1);        n /= 2;        int k = 2, cnt = 0;        while (n) {            cnt = (int)ceil(double(n) / 2);            ans += k * cnt;            k *= 2;            n /= 2;        }        cout << ans << endl;    }    return 0;}