题目链接
第一问：快速幂
第二问：扩欧解线性同余方程
第三问：(BSGS)
三个模板
#include <cstdio>
#include <cmath>
#include <map>
using namespace std;
typedef long long ll;
int a, b, p;
ll x, y, z;
ll exgcd(ll a, ll b, ll &x, ll &y){
if(!b){
x = 1; y = 0; return a;
}
ll d = exgcd(b, a % b, x, y);
ll z = x; x = y; y = z - a / b * y;
return d;
}
int fast_pow(int n, int k){ //n^k%p
int ans = 1;
while(k){
if(k & 1) ans = (ll)ans * n % p;
n = (ll)n * n % p;
k >>= 1;
}
return ans;
}
ll BSGS(){ //a^x≡b(mod p)
map <ll, ll> hash; hash.clear();
int t = ceil(sqrt(p)), val = b, j = 1;
for(int i = 0; i < t; ++i){
hash[val] = i;
val = (ll)val * a % p;
}
a = fast_pow(a, t);
if(!a) return !b ? 1 : -1;
for(int i = 0; i <= t; ++i){
int k = hash.find(j) == hash.end() ? -1 : hash[j];
if(k >= 0 && (i * t - k) >= 0) return (ll)i * t - k;
j = (ll)j * a % p;
}
return -1;
}
int T, K;
int main(){
scanf("%d%d", &T, &K);
if(K == 3)
while(T--){
scanf("%d%d%d", &a, &b, &p);
int ans = BSGS();
ans == -1 ? puts("Orz, I cannot find x!") : printf("%d
", ans);
}
else if(K == 1)
while(T--){
scanf("%d%d%d", &a, &b, &p);
printf("%d
", fast_pow(a, b));
}
else
while(T--){
scanf("%d%d%d", &a, &b, &p);
if(b % (z = exgcd(a, p, x, y))) puts("Orz, I cannot find x!");
else printf("%d
", ((ll)x * b / z % p + p) % p);
}
return 0;
}