先上题目:

C - 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   规律题,如果熟悉树状数组的话说不定可以直接得到答案,如果不熟悉的话可以打表看一下,可以发现前几项lowbit大概是1,2,1,4,1,2,1,8,······发现奇数项都是1,这是lowbit运算导致的,然后偶数项提出来,如果都除以2,发现就会变成1,2,1,4,···然后就有感觉是迭代了,最终得到公式
  ans(x)=2*ans(x/2)+n/2+(n&1) 上代码:
 #include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#define lowbit(x) (x & (-x))
#define MAX 100002
#define LL long long
using namespace std; LL solve(int n){
if(n==) return ;
LL ans = *solve(n>>) + n/;
if(n&) ans++;
return ans;
} int main()
{
int n;
while(scanf("%d",&n)!=EOF){
printf("%lld\n",solve(n));
}
return ;
}

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