ACdream 1154 Lowbit Sum(数学:推理)

Lowbit Sum

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

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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

Source

dream

Manager

scaucontest

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明明是一道很水的题...AC率都接近50%了

我想了好久才想出来

感觉也是因为刷网页刷的不在状态了

以后群赛等挂完了再做...不然全忙着刷网页了

给定一个n

考虑n二进制位从最低位到最高位依次为1的情况

代码如下:

#include <cstdio>#include <cstdlib>#include <bitset>#include <iostream>#include <algorithm>#define LL long longusing namespace std;LL n;int main(void) {    while(cin >> n) {        LL i = 4;        LL tmp = (n+1)/2;        while((n+i/2)/2 > 0) {            tmp += ((n+i/2)/i)*(i/2);            i *= 2;        }        cout << tmp << endl;    }}