#50: Luogu 2485 模板
Posted shandongs1
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$des$
1、给定y、z、p,计算y^z mod p 的值;
2、给定y、z、p,计算满足xy ≡z(mod p)的最小非负整数x;
3、给定y、z、p,计算满足y^x ≡z(mod p)的最小非负整数x。
$sol$
模板+模板+模板
#include <bits/stdc++.h> using namespace std; #define LL long long LL n, k; LL Ksm(LL a, LL b, LL p) { LL ret = 1; while(b) { if(b & 1) ret = ret * a % p; a = a * a % p; b >>= 1; } return ret; } void Work1() { for(int i = 1; i <= n; i ++) { LL y, z, p; cin >> y >> z >> p; cout << Ksm(y, z, p) << " "; } } LL Exgcd(LL a, LL b, LL &x, LL &y) { if(b == 0) { x = 1, y = 0; return a; } LL g = Exgcd(b, a % b, x, y); LL tmpx = x; x = y; y = tmpx - a / b * y; return g; } void Work2() { for(int i = 1; i <= n; i ++) { LL y, z, p; cin >> y >> z >> p; LL x, k; LL gcd = Exgcd(y, p, x, k); if(z % gcd) { puts("Orz, I cannot find x!"); continue; } LL r = z / gcd, d = p / gcd; x *= r; x = ((x % d) + d) % d; cout << x << " "; } } map <LL, int> Map; void Work3() { for(int i = 1; i <= n; i ++) { LL y, z, p; cin >> y >> z >> p; // y ^ x = z (mod p) LL m = sqrt(p); if(m * m != p) m ++; // y ^ {im} = z * y ^ j (mod p) if(!(y % p)) { puts("Orz, I cannot find x!"); continue; } Map.clear(); Map[z % p] = 0; LL ans = z % p; for(int i = 1; i <= m; i ++) { ans = (ans * y) % p; Map[ans] = i; } LL f = Ksm(y, m, p); bool flag = 1; ans = 1; for(int i = 1; i <= m; i ++) { ans = ans * f % p; if(Map[ans]) { LL O = i * m - Map[ans]; O = (O % p + p) % p; cout << O << " "; flag = 0; break; } } if(flag) { puts("Orz, I cannot find x!"); } } } int main() { cin >> n >> k; k == 1 ? Work1() : (k == 2 ? Work2() : Work3()); return 0; }
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