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==)return ;
return ((x+)>>)+(solve(x/)<<);
}
int main()
{
int n;
while(~scanf("%d",&n))
{
printf("%lld\n",solve(n));
}
}