CF 1362C Johnny and Another Rating Drop

传送门

题目：给定一个数n，问（0~n）相邻两个数之间二进制位不同个数的总和。

思路：看出规律，把n转化为二进制，如果该二进制位处于第x位且为1，则它的贡献为2^(x) - 1，累加所有贡献即可。

#include<iostream>
#include<string>
#include<vector>
#include<cstdio>

#define ll long long
#define pb push_back

using namespace std;

const int N = 1e6 + 10;

void solve()
{
    int T;
    cin >> T;
    while(T--){
        ll n, dif;
        cin >> n;

        dif = 0;
        for(int i = 0; i <= 61; ++i){
            ll x = (((n >> i) & 1) << (i + 1));
            dif += x - (x != 0);
        }

        //cout << "dif = " << dif << endl;
        cout << dif << endl;
    }
}

int main() {

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    solve();
    //cout << "ok" << endl;
    return 0;
}