【SDOI2011】计算器
Description
给定$y, z, p$，求$x=y^{z} \mod p$或$xy\equiv z \pmod p$或$y^x\equiv z\pmod p$中x的值
Solution
第一个式子我们可以直接用快速幂求解答案。时间复杂度$O(log_{2}z)$
第二个式子我们可以变形为$xy+Yp=z$，然后用扩展欧几里得算法求出$xy+Yp=gcd(y,p)$的解，然后利用裴蜀定理判断是否有解即可。
第三个式子我们可以应用BSGS算法求解。详细的解释在这里
Code
1 #include <cstdio> 2 #include <map> 3 #include <cmath> 4 using namespace std; 5 typedef long long ll; 6 ll qpow(ll a, ll b, ll p) { 7 ll ret = 1; 8 while (b) { 9 if (b & 1) ret = (ret * a) % p; 10 a = a * a % p; 11 b >>= 1; 12 } 13 return ret % p; 14 } 15 ll exgcd(ll a, ll b, ll &x, ll &y) { 16 if (!b) { 17 x = 1, y = 0; 18 return a; 19 } 20 ll gcd = exgcd(b, a % b, x, y); 21 ll x2 = x, y2 = y; 22 x = y2; 23 y = x2 - (a / b) * y2; 24 return gcd; 25 } 26 map <ll, ll> h; 27 ll BSGS(ll a, ll b, ll p) { 28 h.clear(); 29 b %= p; 30 ll t = (ll)sqrt(p) + 1; 31 for (register int j = 0; j < t; ++j) { 32 ll val = (ll)b * qpow(a, j, p) % p; 33 h[val] = j; 34 } 35 a = qpow(a, t, p); 36 if (a == 0) return b == 0 ? 1 : -1; 37 for (register int i = 0; i <= t; ++i) { 38 ll val = qpow(a, i, p); 39 ll j = h.find(val) == h.end() ? -1 : h[val]; 40 if (j >= 0 && i * t - j >= 0) return i * t - j; 41 } 42 return -1; 43 } 44 int n, k; 45 int main() { 46 ll y = 0, z = 0, p = 0; 47 scanf("%d%d", &n, &k); 48 if (k == 1) 49 while (n--) { 50 scanf("%lld%lld%lld", &y, &z, &p); 51 printf("%lld\n", qpow(y, z, p)); 52 } 53 else if (k == 2) 54 while (n--) { 55 scanf("%lld%lld%lld", &y, &z, &p); 56 ll x = 0, yy = 0; 57 ll ret = exgcd(y, p, x, yy); 58 if (z % ret != 0) {puts("Orz, I cannot find x!"); continue ;} 59 ll now = p / ret; 60 while (x < 0) x += now; 61 printf("%lld\n", (x * (z / ret) % now) + now % now); 62 } 63 else 64 while (n--) { 65 scanf("%lld%lld%lld", &y, &z, &p); 66 ll ret = BSGS(y, z, p); 67 if (ret == -1) puts("Orz, I cannot find x!"); 68 else printf("%lld\n", ret); 69 } 70 return 0; 71 }
