[CF1182F]Maximum Sine

题意：(f(x) = ext{abs}( ext{sin}(frac{p}{q} pi x)))，给定(a,b,p,q)，求(xin[a,b])最大的(f(x))。

题解：div2都这么仙了吗。。。

根据高中数学知识可以推出要求的就是使得(frac{px mod q}{q})最接近(frac12)的(x)，也就是(px mod q)最接近(frac q2)。

有一个结论：([px mod q in [l,r]] = lfloorfrac{px-l}{q} floor - lfloorfrac{px-r-1}{q} floor)。考虑二分(px mod q)与(frac q2)的距离，对于一个(mid)，相当于要求是否存在(x)使得(px mod q in [l=frac q2-mid,r=frac q2+mid])。根据上述结论，这等价于(sum_{x=a}^blfloorfrac{px-l}{q} floor - lfloorfrac{px-r-1}{q} floor)是否(>0)。这个式子是可以类欧求的。

求出最小距离以后，用( ext{exgcd})还原(x)即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll Inf = 1e18;

int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while((ch < '0' || ch > '9') && ch != '-') {
    ch = getchar();
  }
  if(ch == '-') {
    o = -1, ch = getchar();
  }
  while(ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0', ch = getchar();
  }
  return x * o;
}

ll solve(ll n, ll a, ll b, ll c) {
  if(n < 0) {
    return 0;
  }
  if(!n) {
    return b / c;
  }
  if(a >= c || b >= c) {
    return solve(n, a % c, b % c, c) + n * (n + 1) / 2 * (a / c) + (n + 1) * (b / c);
  }
  ll m = (a * n + b) / c;
  return m * n - solve(m - 1, c, c - b - 1, a);
}

ll solve(ll l, ll r, ll a, ll b, ll c) {
  return solve(r, a, b, c) - solve(l - 1, a, b, c);
}

void exgcd(ll a, ll b, ll &x, ll &y) {
  if(!b) {
    x = 1, y = 0;
    return;
  }
  exgcd(b, a % b, y, x), y -= a / b * x;
}

ll a, b, p, q;

ll solve(ll p, ll q, ll t) {
  ll gg = __gcd(p, q);
  if(t % gg != 0) {
    return Inf;
  }
  p /= gg, q /= gg, t /= gg;
  ll x, y;
  exgcd(p, q, x, y);
  x *= t, y *= t;
  ll k = (a - x) / q;
  x += k * q;
  while(x >= a) {
    x -= q;
  }
  while(x < a) {
    x += q;
  }
  return x;
}

int main() {
#ifndef ONLINE_JUDGE
  freopen("a.in", "r", stdin);
  freopen("a.out", "w", stdout);
#endif
  int T = gi();
  while(T--) {
    a = gi(), b = gi(), p = gi() << 1, q = gi() << 1;
    ll l = 0, w = q / 2, r = w;
    while(l < r) {
      ll mid = (l + r) >> 1;
      ll L = w - mid, R = w + mid;
      if(solve(a, b, p, q - L, q) - solve(a, b, p, q - R - 1, q)) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    cout << min(solve(p, q, w - l), solve(p, q, w + l)) << endl;
  }
  return 0;
}