[Educational Codeforces Round 16]D. Two Arithmetic Progressions

[Educational Codeforces Round 16]D. Two Arithmetic Progressions

试题描述

You are given two arithmetic progressions: a₁k + b₁ and a₂l + b₂. Find the number of integers x such that L ≤ x ≤ R andx = a₁k' + b₁ = a₂l' + b₂, for some integers k', l' ≥ 0.

输入

The first line contains integer n (1 ≤ n ≤ 3·10⁵) — the number of points on the line.

The second line contains n integers xi ( - 10⁹ ≤ xi ≤ 10⁹) — the coordinates of the given n points.

输出

Print the only integer x — the position of the optimal point on the line. If there are several optimal points print the position of the leftmost one. It is guaranteed that the answer is always the integer.

输入示例

输出示例

数据规模及约定

解一下不定方程 a₁k + b₁ = a₂l + b₂，设 k mod lcm(a₁, a₂) / a₁ 的值是 t，设 lcm(a₁, a₂) / a₁ = A，那么 k 可以写成 q·A + t 这个样子，那么显然 A 是有上下界的，我们二分到这个上下界，做个差就是答案了。

一上午就调它了。。。woc cf 数据太强了

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

#define MAXN 1000+10

#define oo 4000000000ll
#define LL long long
#define LD long double
LL a1, b1, a2, b2, L, R;

LL gcd(LL a, LL b, LL& x, LL& y) {
	if(b == 0){ x = 1, y = 0; return a; }
	LL d = gcd(b, a % b, y, x); y -= (a / b) * x;
	return d;
}

LL gcd(LL a, LL b) { return b == 0 ? a : gcd(b, a % b); }

int main() {
	cin >> a1 >> b1 >> a2 >> b2 >> L >> R;

	LL k, t;
	LL d = gcd(a1, a2, k, t);
	if((b2 - b1) % d != 0) return puts("0"), 0;
	k *= (b2 - b1) / d; t *= (b2 - b1) / d;
	LL A2 = a2 / gcd(a1, a2);
	LL mod = (k % A2 + A2) % A2, al, ar;
//	printf("%lld %lld\n", k, mod);
	LL l, r; l = -oo - 1; r = oo + 1;
//	printf("%lld %lld\n", l, r);
	while(l < r) {
		LL mid = l + (r - l) / 2;
		LL lsid = (L - b1) % a1 != 0 ? (L - b1) / a1 + (L - b1 > 0 ? 1 : 0) : (L - b1) / a1,
		   rsid = (R - b1) % a1 != 0 ? (R - b1) / a1 + (R - b1 > 0 ? 0 : -1) : (R - b1) / a1,
		   x = mid * A2 + mod;
		LL l2 = (L - b2) % a2 != 0 ? (L - b2) / a2 + (L - b2 > 0 ? 1 : 0) : (L - b2) / a2,
		   r2 = (R - b2) % a2 != 0 ? (R - b2) / a2 + (R - b2 > 0 ? 0 : -1) : (R - b2) / a2,
		   y = ((LD)a1 * x - b2 + b1) / a2;
//		printf("%lld %lld %lld %lld %lld [%lld, %lld]\n", lsid, l2, mid, y, x, l, r);
		if(lsid <= x && l2 <= y && x >= 0 && y >= 0) r = mid;
		else l = mid + 1;
	}
	al = l;
	l = -oo - 1; r = oo + 1;
//	printf("%lld %lld\n", l, r);1 -2000000000 2 2000000000 -2000000000 2000000000
	while(l < r - 1) {
		LL mid = l + (r - l) / 2;
		LL lsid = (L - b1) % a1 != 0 ? (L - b1) / a1 + (L - b1 > 0 ? 1 : 0) : (L - b1) / a1,
		   rsid = (R - b1) % a1 != 0 ? (R - b1) / a1 + (R - b1 > 0 ? 0 : -1) : (R - b1) / a1,
		   x = mid * A2 + mod;
		LL l2 = (L - b2) % a2 != 0 ? (L - b2) / a2 + (L - b2 > 0 ? 1 : 0) : (L - b2) / a2,
		   r2 = (R - b2) % a2 != 0 ? (R - b2) / a2 + (R - b2 > 0 ? 0 : -1) : (R - b2) / a2,
		   y = ((LD)a1 * x - b2 + b1) / a2;
//		printf("%lld %lld %lld %lld %lld [%lld, %lld]\n", mid, x, y, rsid, r2, l, r);
		if(x <= rsid && y <= r2) l = mid;
		else r = mid;
	}
	ar = l;
//	printf("%lld %lld\n", al, ar);

	LL mid = l + (r - l) / 2;
	LL lsid = (L - b1) % a1 != 0 ? (L - b1) / a1 + (L - b1 > 0 ? 1 : 0) : (L - b1) / a1,
	   rsid = (R - b1) % a1 != 0 ? (R - b1) / a1 + (R - b1 > 0 ? 0 : -1) : (R - b1) / a1,
	   x = mid * A2 + mod;
	LL l2 = (L - b2) % a2 != 0 ? (L - b2) / a2 + (L - b2 > 0 ? 1 : 0) : (L - b2) / a2,
	   r2 = (R - b2) % a2 != 0 ? (R - b2) / a2 + (R - b2 > 0 ? 0 : -1) : (R - b2) / a2,
	   y = ((LD)a1 * x - b2 + b1) / a2;
	if(lsid <= x && x <= rsid && l2 <= y && y <= r2 && x >= 0 && y >= 0 && al <= ar)
		cout << ar - al + 1 << endl;
	else puts("0");

	return 0;
}