[SDOI2011]计算器

嘟嘟嘟

黄题 + 绿题 + 蓝题 = 紫题……
对于询问1，直接快速幂。
对于询问2，\(exgcd\)。
对于询问3,\(bsgs\)，但要特判一下\(a \ \ mod \ \ c = 0\)且 \(b \ \ mod \ \ c \neq 0\)的时候应该无解。
\(bsgs\)不懂的可以看我的这篇博客

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
const int base = 999979;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

ll y, z, p;

void work1(ll a, ll b, ll mod)
{
  a %= mod; ll ret = 1;
  for(; b; b >>= 1, a = a * a % mod)
    if(b & 1) ret = ret * a % mod;
  write(ret), enter;
}

void exgcd(ll a, ll b, ll& d, ll& x, ll& y)
{
  if(!b) d = a, x = 1, y = 0;
  else exgcd(b, a % b, d, y, x), y -= x * (a / b);
}
void work2(ll a, ll c, ll b)
{
  ll x, y, d;
  exgcd(a, b, d, x, y);
  if(c % d) {puts("Orz, I cannot find x!"); return;}
  ll t = b / d;
  write((x * (c / d) % t + t) % t), enter;
}

struct Hash
{
  int nxt; ll to; int w;
}e[maxn];
int head[base], hcnt = 0;
int st[maxn], top = 0;
void init()
{
  hcnt = 0;
  while(top) head[st[top--]] = 0;
}
void insert(ll x, int y)
{
  int h = x % base;
  if(!head[h]) st[++top] = h;
  e[++hcnt] = (Hash){head[h], x, y};
  head[h] = hcnt;
}
int query(ll x)
{
  int h = x % base;
  for(int i = head[h]; i; i = e[i].nxt)
    if(e[i].to == x) return e[i].w;
  return -1;
}

int bsgs(ll a, ll b, ll c)
{
  if(a % c == 0 && b % c) return -1;
  init();
  int s = sqrt(c);
  ll p = 1;
  for(int i = 0; i < s; ++i)
    {
      if(p == b) return i;
      insert(p * b % c, i);
      p = p * a % c;
    }
  ll q = p;
  for(int i = 1; i <= s; ++i)
    {
      int t = query(q);
      if(t != -1) return i * s - t;
      q = q * p % c;
    }
  return -1;
}
void work3(ll a, ll b, ll c)
{
  b %= c;
  ll ans = bsgs(a, b, c);
  if(ans == -1) puts("Orz, I cannot find x!");
  else write(ans), enter;
}

int main()
{
  int T = read(), K = read();
  while(T--)
    {
      y = read(); z = read(); p = read();
      if(K == 1) work1(y, z, p);
      else if(K == 2) work2(y, z, p);
      else work3(y, z, p);
    }
  return 0;
}