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C. Interesting Sequence

By xiaruize

思路

手模一下样例一

10=(1010)210 = (1010)_2 , 发现需要将 212^1 位的 11 去掉

此时，就需要与一个 212^1 位为 00 的数, 同时其它位不能使 1010 改变, 且这个数 ≥10\geq 10

那么, 这个数的最小值显然为 (1100)2=12(1100)_2=12

总结上面的过程, 得到以下这些结论

  1. n&x=xn \And x=x , 否则是 -1

  2. 考虑 n−xn - x 的值的二进制表示 (即 nn 中需要改为 00 的二进制位的值), 发现所有的 11 必须是 nn 中最低且连续的, 否则 -1

  3. 设 n−xn-x 中最高位的 11 在 2k2^k 位上, 那么 2k+12^{k+1} 位在 nn 中一定为 00 , 否则 -1

  4. 否则答案为 2k+1−n+x2^{k+1}-n+x

samples

  1. n=11,m=4n=11,m=4
  2. n=20,m=16n=20,m=16
  3. n=11,m=9n=11,m=9

如果不能理解结论，建议手模一遍上述样例

Code

cpp
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/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int n, x;
int t;
// bool en;

void solve()
{
    cin >> n >> x;
    if (x > n)
    {
        cout << "-1" << endl;
        return;
    }
    if (x == n)
    {
        cout << n << endl;
        return;
    }
    int p = n - x;
    int d = p;
    for (int i = 0; i <= 61; i++)
    {
        if (p < 0)
        {
            cout << "-1" << endl;
            return;
        }
        if (p == 0 && (n & (1ll << i)))
        {
            cout << "-1" << endl;
            return;
        }
        if (p == 0)
        {
            cout << (1ll << i) - d + n << endl;
            return;
        }
        if (n & (1ll << i))
        {
            p -= (1ll << i);
        }
    }
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> t;
    while (t--)
        solve();
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}