2019 Multi-University Training Contest 6 Problem 1008 & HDU6641

                                                    TDL

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0
 

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 nthat (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

 

巧妙的暴力，由于公式的关系可得知i与k值相差不远，我们可以设立一个差值，在差值中遍历i即可。（实测500也可以）

#include 
using namespace std;
#define ll long long
const int MAXN = 1e3 + 7;

ll gcd(ll a, ll b)
{
    return b ? gcd(b, a%b) : a;
}

ll fun(ll n, ll m)
{
    ll i = n;
    while(m)
    {
        i++;
        if(gcd(i, n) == 1) m--;
    }
    return i;
}

int main()
{
    ios::sync_with_stdio(0);
    ll t, k, m;
    int flag;
    cin>>t;
    while(t--)
    {
        cin>>k>>m;
        flag = 1;
        for(ll i = max((ll)1, k-MAXN); i < k+MAXN; i++)
        {
            if(((fun(i, m) - i)^i) == k)
            {
                cout<