Lowbit Sum 规律

Lowbit Sum

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

SubmitStatus

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

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define ll long long
ll solve(int x)
{
    if(x==0)return 0;
    return ((x+1)>>1)+(solve(x/2)<<1);
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        printf("%lld
",solve(n));
    }
}