Ultraviolet A 1426-Discrete Square Roots (number theory)

Ultraviolet A 1426-Discrete Square Roots

Question Link

Question: given X, N, R, requirementsR2 coresX(MOD N) (1 <= r <n). R is a known solution.

Ideas:
R2 coresX(MOD N) =>R2 +K1N=X
OneR!, Bring two ways to subtractR2?R12 =KN=> (R+R1 )(R?R1) =KN
Enumerate A and B so that
A * B = n
(R + r1) is A multiple
(R-r1) is a multiple of B.
In this way, we can launch
AKA?R1 =BKB+R1 =R
=>AKA=BKB+ 2R1
=>AKALimit 2R1 (MOD B)
This is equivalent to finding all the solutions of the linear modulus equation, then finding another solution R, and finally saving all the answers with one set to output

Code:

#include 
 
  #include 
  
   #include 
   
    #include 
    
     using namespace std;long long X, N, R;set
     
       ans;long long exgcd(long long a, long long b, long long &x, long long &y) {    if (!b) {x = 1; y = 0; return a;}    long long d = exgcd(b, a % b, y, x);    y -= a / b * x;    return d;}void mod_line(long long a, long long b, long long n) {    long long x, y;    long long d = exgcd(a, n, x, y);    if (b % d) return;    x = x * (b / d);    x = (x % (n / d) + (n / d)) % (n / d);    long long a0 = x * a - b / 2;    long long k = a * n / d;    for (long long tmp = a0; tmp < N; tmp += k) {if (tmp >= 0) ans.insert(tmp);    }}int main() {    int cas = 0;    while (~scanf("%lld%lld%lld", &X, &N, &R) && N) {ans.clear();long long m = (long long)sqrt(N);for (long long i = 1; i <= m; i++) {    if (N % i) continue;    mod_line(i, 2 * R, N / i);    mod_line(N / i, 2 * R, i);}printf("Case %d:", ++cas);for (set
      
       ::iterator it = ans.begin(); it != ans.end(); it++) printf(" %lld", *it);printf("\n"); } return 0;}