2019 Multi-University Training Contest 6 TDL

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 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

题意

f (n, m) 代表比 n 大的第 m 个与 n 互质的数，现已知 (f(n,m)−n)⊕n 的值为 k，求 n 的值

分析

(f(n,m)−n)⊕n=k 可以写成 f(n,m)-n=n⊕k，因为 m 的值为100，所以(f(n,m)−n) 的值比较小，则 n⊕k 的值比较小，说明n 与 k 相接近，所以对 k-512 到 k+512 之间的数进行遍历，找到适合的 n 即是答案。

代码

#include
using namespace std;
int main(){
    int t;
    scanf("%d",&t);
    long long k,i;
    int m;
    while(t--)
	{
        scanf("%lld%d",&k,&m);
        if(k<512)	i=1;
        else 		i=k-512;
        long long ans=0;
        for( ;i