HDU

Problem Description

For a positive integer n, let's denote function f(n,m) as the m-th smallest integer x that x>n and gcd(x,n)=1. For example, f(5,1)=6 and f(5,5)=11.

You are given the value of m and (f(n,m)−n)⊕n, where ``⊕'' denotes the bitwise XOR operation. Please write a program to find the smallest positive integer n that (f(n,m)−n)⊕n=k, or determine it is impossible.

 

Input

The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.

In each test case, there are two integers k,m(1≤k≤1018,1≤m≤100).

 

Output

For each test case, print a single line containing an integer, denoting the smallest n. If there is no solution, output ``-1'' instead.

 

Sample Input

2 3 5 6 100

 

Sample Output

5 -1

 

Source

2019 Multi-University Training Contest 6

题解：

///
///                            _ooOoo_
///                           o8888888o
///                           88" . "88
///                           (| -_- |)
///                           O  =  /O
///                        ____/`---'\____
///                      .'  \|     |//  `.
///                     /  \|||  :  |||//  
///                    /  _||||| -:- |||||-  
///                    |   | \  -  /// |   |
///                    | \_|  ''---/''  |   |
///                      .-\__  `-`  ___/-. /
///                  ___`. .'  /--.--  `. . __
///               ."" '<  `.___\_<|>_/___.'  >'"".
///              | | :  `- \`.;` _ /`;.`/ - ` : | |
///                 `-.   \_ __ /__ _/   .-` /  /
///         ======`-.____`-.___\_____/___.-`____.-'======
///                            `=---='
///        ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
///                      Buddha Bless, No Bug !
///
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <vector>
using namespace std;
#define MAXN 100010
#define ll long long

int t, m;
ll k, ans_n;

ll cal(ll n, int m)
{
    if(n < 1)
        return 0;
    for(ll i = n + 1;  ; i++)
        if(__gcd(n, i) == 1)
        {
            m--;
            if(m == 0)
                return i - n;/// (i - n) = (f(n, m) - n) = d
        }
}

int main()
{
    scanf("%d", &t);
    while(t--)
    {
        scanf("%lld%d", &k, &m);
        ans_n = -1;
        for(int d = 1; d <= 1000; d++)
        {
            if(cal(k ^ d, m) == d)
            {
                if(ans_n == -1)
                    ans_n = k ^ d;
                else if(ans_n > (d ^ k))/// ^ 运算的有优先度小于 <  >  ==  !=
                    ans_n = k ^ d;
            }
        }
        printf("%lld
", ans_n);
    }
    return 0;
}