2018 CCPC-Final K - Mr. Panda and Kakin

\(c\equiv FLAG^{2^{30}+3} ~~ (mod~~n)\)
给定\(c\)和\(n\)，\(n\)为两接近的素数乘积，求\(FLAG\)

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\(RSA\)：

• 寻找两个质数\(p\)，\(q\)，令\(n=p*q\)
• 计算欧拉函数\(\varphi(n)=(p-1)*(q-1)\)
• 寻找公钥\(e\)，满足\(1< e < \varphi(n)\)且\(e\)与\(\varphi(n)\)互质
• 计算私钥\(d\)，\(e*d\equiv 1 ~~ (mod ~~ \varphi(n))\)

若\(c\)为密文，\(m\)为明文
加密：\(c=m^e ~ mod ~ n\)
解密：\(m=c^d ~ mod ~ n\)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll mul(ll u, ll v, ll p) {
    return (u * v - ll((long double)u * v / p) * p + p) % p;
}
ll qpow(ll x, ll y, ll mod) {
    ll ans = 1;
    for (; y; y >>= 1, x = mul(x, x, mod))
        if (y & 1) ans = mul(ans, x, mod);
    return ans;
}
ll extend_gcd(ll a, ll b, ll &x, ll &y) {
    if (a == 0 && b == 0) return -1;
    if (!b) {
        x = 1, y = 0;
        return a;
    }
    ll d = extend_gcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}
ll inv(ll a, ll n) {
    ll x, y;
    ll d = extend_gcd(a, n, x, y);
    if (d == 1) return (x % n + n) % n;
    return -1;
}
int main() {
    int t;
    ll n, c;
    scanf("%d", &t);
    for (int _ = 1; _ <= t; _++) {
        scanf("%lld%lld", &n, &c);
        ll qn = sqrt(n), q, p;
        while (1) {
            if (n % qn == 0) {
                q = qn;
                p = n / q;
                break;
            }
            qn--;
        }
        ll phi = (q - 1) * (p - 1);
        ll d = inv(1073741827ll, phi);
        printf("Case %d: %lld\n", _, qpow(c, d, n));
    }
    return 0;
}