Mr. Panda and Kakin

給定$n,c$，要我們找到$n$是兩個相鄰質數的乘積，要我們找到$x$，滿足$x^{2^{30}+3}\equiv c(\mathrm{m}\mathrm{o}\mathrm{d}n)$，$10^{10}\le n\le 10^{18},0<c<n$，

考慮得到$2^{30}+3$模$?(n)$下的逆元，為$inv=(2^{30}+3{)}^{?(n)?1}$，則有$c^{inv}\equiv x^{(2^{30}+3)inv}\equiv x(\mathrm{m}\mathrm{o}\mathrm{d}n)$，使用$O(1)$快速乘即可。

#include <bits/stdc++.h>using namespace std;long long n, c, phi, inv;inline long long mul(long long x, long long y, long long mod) {return (x * y - (long long)((long double)x / mod * y) * mod + mod) % mod; }long long exgcd(long long a, long long b, long long & x, long long & y) {if(!b) {x = 1, y = 0;sreturn a;}long long gcd = exgcd(b, a % b, x, y);long long temp = x;x = y;y = temp - a / b * y;return gcd; }long long quick_pow(long long a, long long n, long long mod) {long long ans = 1;while (n) {if (n & 1) {ans = mul(ans, a, mod);}a = mul(a, a, mod);n >>= 1;}return ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);int T, cas = 0;scanf("%d", &T);while (T--) {scanf("%lld %lld", &n, &c);long long p = sqrt(n), q;while (true) {if (n % p == 0) {break;}p--;}q = n / p;phi = (p - 1) * (q - 1);exgcd((1ll << 30) + 3, phi, inv, p);inv = ((inv % phi) + phi) % phi;printf("Case %d: %lld\n", ++cas, quick_pow(c, inv, n));}return 0; }

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