题目链接：uva 1426 - Discrete Square Roots

题目大意：给出X,N,R,求出所有满足的r，使得r2≡x%N,并且R是一个其中的解。

解题思路：

• R2?r2=k?N
• (R?r)(R+r)=k?N
• => aA=(R+r),bB=(R?r)，A，B为N的因子
  所以枚举A，B，就有r=R?aA=bB?R
• aA+bB=2?R
  拓展欧几里得求解，将所有满足的解放入set中自动去重。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <set>
#include <algorithm>

using namespace std;
typedef long long ll;

ll X, N, R;
set<ll> ans;

void exgcd (ll a, ll b, ll& d, ll& x, ll& y) {
    if (b == 0) {
        d = a;
        x = 1;
        y = 0;
    } else {
        exgcd (b, a%b, d, y, x);
        y -= (a/b) * x;
    }
}

void solve (ll A, ll B, ll C) {
    ll d, x, y;
    exgcd(A, B, d, x, y);

    if (C % d)
        return;

    x = x * C / d;
    x = (x % (B/d) + (B/d))%(B/d);
    ll s = x * A - C / 2;
    ll k = A * B / d;
    for (ll i = s; i < N; i += k)
        if (i >= 0)
            ans.insert(i);

}

int main () {
    int cas = 1;
    while (scanf("%lld%lld%lld", &X, &N, &R) == 3 && X + N + R) {
        ans.clear();
        ll m = sqrt(N+0.5);
        for (ll i = 1; i <= m; i++) {
            if (N%i)
                continue;
            solve(i, N/i, 2*R);
            solve(N/i, i, 2*R);
        }

        printf("Case %d:", cas++);
        for (set<ll>::iterator i = ans.begin(); i != ans.end(); i++)
            printf(" %lld", *i);
        printf("\n");
    }
    return 0;
}