2019HDU more than school sixth 6641 TDL

I, entitled

　　TDL

Second, analysis　　

　　Meaning of the questions is to find a formula for $ n $ title meet, can not find it outputs $ -1 $.

　　For $$ {(f (n, m) - n)} \ oplus {n} = k $$

　　Can be converted to look into $ (f (n, m) - n) = {k} \ oplus {n} $, while for $ f (n, m) - n $ may play table look, understood it according to their mass number density small.

　　Since the XOR is equivalent to an addition of the carry and because $ F ( n- , m ) - n- $ small, very close to the equivalent of $ n-$ $ k $.

　　Enumeration $ n $, since there is no time limit card, the enumeration range very relaxed.

Three, AC codes

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 #define ll long long
 5 #define Min(a,b) ((a)>(b)?(b):(a))
 6 #define Max(a,b) ((a)>(b)?(a):(b))
 7 
 8 ll gcd(ll a, ll b)
 9 {
10     return b == 0 ? a : gcd(b, a % b);
11 }
12 
13 ll f(ll n, int m)
14 {
15     int cnt = 0;
16     for(ll i = n + 1;;i++)
17     {
18         if(gcd(i, n) == 1)
19         {
20             cnt++;
21         }
22         if(cnt == m)
23             return i;
24     }
25 }
26 
27 int main()
28 {
29     int T;
30     scanf("%d", &T);
31     while(T--)
32     {
33         int m;
34         ll k, n;
35         scanf("%lld %d", &k, &m);
36         for(n = Max(1, k - 1000); n < k + 1000; n++)
37         {
38             if( f(n, m) - n == (k^n) )
39                 break;
40         }
41         if(n == k + 1000)
42             puts("-1");
43         else
44             printf("%lld\n", n);
45     }
46     return 0;
47 }