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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#877882 | #9434. Italian Cuisine | 2317663977 | AC ✓ | 18ms | 7020kb | C++23 | 21.9kb | 2025-02-01 11:44:25 | 2025-02-01 11:44:26 |
Judging History
answer
#include <iostream>
using namespace std;
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <cmath>
#include <cstring>
#include <iomanip>
namespace Geo {
template<typename T>
class Point;
template<typename T>
class Line;
template<typename T>
class Polygon;
template<typename T>
class Circle;
template<typename T>
using Vector = Point<T>;
template<typename T>
using Segment = Line<T>;
template<typename T>
using PointSet = Polygon<T>;
const double eps = 1e-9;
const double PI = acos(-1.0);
// 浮点数 x 的符号
template<typename T>
int sgn(T x) {
if (fabs(x) < eps) return 0;
return x > 0 ? 1 : -1;
}
// 比较两个浮点数
template<typename T>
int cmp(T x, T y) {
if (fabs(x - y) < eps)return 0; // x == y,返回 0
else return x < y ? -1 : 1; // x < y,返回 -1; x > y,返回 1
}
double radians(double degrees) {
return degrees * PI / 180.0;
}
// 两点距离
template<typename T>
double dist(const Point<T>& A, const Point<T>& B) {
return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}
// 点积
template<typename T>
T dot(const Vector<T>& A, const Vector<T>& B) {
// a · b = |a| |b| cos
// 可用于判断两向量夹角
return A.x * B.x + A.y * B.y;
}
// 叉积
template<typename T>
T cross(const Vector<T>& A, const Vector<T>& B) {
// a · b = |a| |b| sin
// 可以判断两向量的相对方向
// 也能算两向量形成的平行四边形面积
return A.x * B.y - A.y * B.x;
}
// 向量长度
template<typename T>
T len(const Vector<T>& A) {
return sqrt(dot(A, A));
}
// 向量长度的平方
template<typename T>
T len2(const Vector<T>& A) {
return dot(A, A);
}
// 两向量夹角
template<typename T>
double angle(const Vector<T>& A, const Vector<T>& B) {
return acos(dot(A, B) / len(A) / len(B));
}
// 计算两向量构成的平行四边形有向面积
// 三个点A、B、C,以A为公共点,得到2个向量B-A和C-A,它们构成的平行四边形
template<typename T>
T area_parallelogram(const Point<T>& A, const Point<T>& B, const Point<T>& C) {
return abs(cross(B - A, C - A));
}
// 计算两向量构成的平行四边形面积
// 两个有公共点的向量 A B 构成的平行四边形
template<typename T>
T area_parallelogram(const Vector<T>& A, const Vector<T>& B) {
return abs(cross(A, B));
}
// 计算两向量构成的三角形面积
// 三个点A、B、C,以A为公共点,得到2个向量B-A和C-A,它们构成的三角形
template<typename T>
T area_triangle(const Point<T>& A, const Point<T>& B, const Point<T>& C) {
return abs(cross(B - A, C - A)) / 2.0;
}
// 计算两向量构成的三角形有向面积
// 两个有公共点的向量 A B 构成的三角形
template<typename T>
T area_triangle(const Vector<T>& A, const Vector<T>& B) {
return abs(cross(A, B)) / 2.0;
}
// 向量旋转
/*
特殊情况是旋转90度:
逆时针旋转90度:Rotate(A, pi/2),返回Vector(-A.y, A.x);
顺时针旋转90度:Rotate(A, -pi/2),返回Vector(A.y, - A.x)。
*/
template<typename T>
Vector<T> rotate(const Vector<T>& A, double rad) {
return Vector<T>(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}
// 有时需要求单位法向量,即逆时针转90度,然后取单位值。
template<typename T>
Vector<T> normal(const Vector<T>& A) {
return Vector<T>(-A.y / len(A), A.x / len(A));
}
// 两个向量是否平行或重合
template<typename T>
bool parallel(const Vector<T>& A, const Vector<T>& B) {
return sgn(cross(A, B)) == 0; // 返回true表示平行或重合
}
// 点和直线的位置关系
template<typename T>
int point_line_relation(const Point<T>& p, const Line<T>& v) {
int c = sgn(cross(p - v.p1, v.p2 - v.p1));
if (c < 0)return 1; // 1 :p在v的左边
if (c > 0)return 2; // 2 :p在v的右边
return 0; // 0 :p在v上
}
// 点和线段的位置关系
template<typename T>
bool point_segment_relation(const Point<T>& p, const Line<T>& v) {
// 0 点不在线段v上
// 1 点在线段v上
// 前者为 True 说明 p 和 线段 v 的一个端点连边,和 v 本身的夹角为 0,即 p 在 直线 v 上
// 后者为 True 说明 p 和两端点形成平角,也就是说 p 在两端点之间
return sgn(cross(p - v.p1, v.p2 - v.p1)) == 0 && sgn(dot(p - v.p1, p - v.p2)) <= 0;
}
// 点到直线的距离
template<typename T>
double point_line_dis(const Point<T>& p, const Line<T>& v) {
// 实际上是算了 p 和 v 的一个端点连边,然后和 v 形成的平行四边形的面积,除底得到
return fabs(cross(p - v.p1, v.p2 - v.p1)) / dist(v.p1, v.p2);
}
// 点在直线上的投影
template<typename T>
Point<T> point_line_proj(const Point<T>& p, const Line<T>& v) {
double k = dot(v.p2 - v.p1, p - v.p1) / len2(v.p2 - v.p1);
return v.p1 + (v.p2 - v.p1) * k;
}
// 点关于直线的对称点
template<typename T>
Point<T> point_line_symmetry(const Point<T>& p, const Line<T>& v) {
Point<T> q = point_line_proj(p, v);
return Point<T>(2 * q.x - p.x, 2 * q.y - p.y);
}
// 点到线段的距离
template<typename T>
double point_segment_dis(const Point<T>& p, const Segment<T>& v) {
// 先检查点 p 到线段 v 的投影是否在线段 v 上
// 如果在,就直接返回点 p 到直线 v 距离
// 如果不在,就返回线段 v 两端点里 p
// p 先和 v 的两端点比较,看看是否两向量夹角 > 90
if (sgn(dot(p - v.p1, v.p2 - v.p1)) < 0 || sgn(dot(p - v.p2, v.p1 - v.p2)) < 0)
return min(dist(p, v.p1), dist(p, v.p2)); // 点的投影不在线段上
return point_line_dis(p, v); // 点的投影在线段上
}
// 叉积判断两条向量的位置关系,AB * AC,两向量共点
template<typename T>
int vector_vector_relation(const Vector<T>& v1, const Vector<T>& v2) {
int sign = sgn(cross(v1, v2));
if (sign == 0)return 0; // 共线
if (sign > 0)return 1; // v2 在 v1 的逆时针方向
if (sign < 0)return 2; // v2 在 v1 的顺时针方向
}
// 叉积判断两条直线的位置关系
template<typename T>
int line_line_relation(const Line<T>& v1, const Line<T>& v2) {
if (sgn(cross(v1.p2 - v1.p1, v2.p2 - v2.p1)) == 0) {
if (point_line_relation(v1.p1, v2) == 0) return 1; // 1: 重合
else return 0; // 0: 平行
}
return 2; // 2: 相交
}
// 两向量夹角类型
template<typename T>
int vector_vector_angle_type(const Vector<T>& v1, const Vector<T>& v2) {
// 0:夹角度为 0
// 1:夹角为锐角
// 2:夹角为钝角
// 3:夹角为平角,即方向相反
auto _dot = dot(v1, v2);
if (vector_vector_relation(v1, v2) == 0) { // 两向量共线
if (sgn(_dot) > 0)return 0;
else return 3;
}
else {
if (sgn(_dot) > 0)return 1;
else return 2;
}
}
// 两条直线的交点
template<typename T>
Point<T> line_line_cross_point(const Point<T>& a, const Point<T>& b, const Point<T>& c, const Point<T>& d) {
// Line1 : ab, Line2 : cd
double s1 = cross(b - a, c - a);
double s2 = cross(b - a, d - a); // 叉积有正负
return Point<T>(c.x * s2 - d.x * s1, c.y * s2 - d.y * s1) / (s2 - s1);
}
// 两条直线的交点
template<typename T>
Point<T> line_line_cross_point(const Line<T>& x, const Line<T>& y) {
// Line1 : ab, Line2 : cd
auto a = x.p1;
auto b = x.p2;
auto c = y.p1;
auto d = y.p2;
double s1 = cross(b - a, c - a);
double s2 = cross(b - a, d - a); // 叉积有正负
return Point<T>(c.x * s2 - d.x * s1, c.y * s2 - d.y * s1) / (s2 - s1);
}
// 两个线段是否相交
template<typename T>
bool segment_segment_is_cross(const Point<T>& a, const Point<T>& b, const Point<T>& c, const Point<T>& d) {
// Line1 : ab, Line2 : cd
double c1 = cross(b - a, c - a), c2 = cross(b - a, d - a);
double d1 = cross(d - c, a - c), d2 = cross(d - c, b - c);
return sgn(c1) * sgn(c2) < 0 && sgn(d1) * sgn(d2) < 0; // 1: 相交;0: 不相交
}
// 点和多边形的关系
template<typename T>
int point_polygon_relation(const Point<T>& pt, const Polygon<T>& p) {
// 点pt,多边形 p
int n = p.size();
for (int i = 0; i < n; i++) { // 枚举点
if (p[i] == pt) return 3; // 3:点在多边形的顶点上
}
for (int i = 0; i < n; i++) { // 枚举边
Line<T> v = Line<T>(p[i], p[(i + 1) % n]);
if (point_segment_relation(pt, v)) return 2; // 2:点在多边形的边上
}
// 通过射线法计算点是否在多边形内部 (具体原理可以看书)
int num = 0;
for (int i = 0; i < n; i++) {
int j = (i + 1) % n;
int c = sgn(cross(pt - p[j], p[i] - p[j]));
int u = sgn(p[i].y - pt.y);
int v = sgn(p[j].y - pt.y);
if (c > 0 && u < 0 && v >= 0) num++;
if (c < 0 && u >= 0 && v < 0) num--;
}
return num != 0; // 1:点在内部; 0:点在外部
}
// 计算多边形周长
template<typename T>
T polygon_perimeter(const Polygon<T>& p) {
T ans = 0;
int n = p.size();
for (int i = 0; i < n; i++)
ans += dist(p[i], p[(i + 1) % n]);
return ans;
}
// 多边形的面积
template<typename T>
T polygon_area(const Polygon<T>& p) {
/*
注意面积存在 正负
逆时针遍历点算出来就是正的
顺时针遍历点算出来就是负的
*/
T area = 0;
int n = p.size();
for (int i = 0; i < n; i++)
area += cross(p[i], p[(i + 1) % n]);
return abs(area) / 2.0;
}
// 多边形的重心
template<typename T>
Point<T> polygon_center_point(const Polygon<T>& p) { //求多边形重心
Point<T> ans(0, 0);
int n = p.size();
if (polygon_area(p, n) == 0) return ans;
for (int i = 0; i < n; i++)
ans = ans + (p[i] + p[(i + 1) % n]) * cross(p[i], p[(i + 1) % n]);
return ans / polygon_area(p, n) / 6;
}
/*
求凸包,凸包顶点放在ch中
凸多边形:是指所有内角大小都在 [0, 180] 范围内的简单多边形
凸包:在平面上能包含所有给定点的最小凸多边形叫做凸包
*/
template<typename T>
Polygon<T> convex_hull(vector<Point<T>> p) {
Polygon<T> ch;
if (p.size() == 0 or p.size() == 1 or p.size() == 2) {
for (auto& i : p) ch.push_back(i);
return ch;
}
// Andrew 法:
// 先对所有点排序
// 求上下凸包 (查每个边相较于上一条边的拐弯方向)
// 然后合并
// 最后得到的点是按照原点的逆时针顺序的
int n = p.size();
n = unique(p.begin(), p.end()) - p.begin(); // 去除重复点
ch.resize(2 * n);
sort(p.begin(), p.end()); // 对点排序:按 x 从小到大排序,如果 x 相同,按 y 排序
int v = 0;
// 求下凸包,如果p[i]是右拐弯的,这个点不在凸包上,往回退
for (int i = 0; i < n; i++) {
while (v > 1 && sgn(cross(ch[v - 1] - ch[v - 2], p[i] - ch[v - 1])) <= 0)
v--;
ch[v++] = p[i];
}
// 求上凸包
for (int i = n - 1, j = v; i >= 0; i--) {
while (v > j && sgn(cross(ch[v - 1] - ch[v - 2], p[i] - ch[v - 1])) <= 0)
v--;
ch[v++] = p[i];
}
ch.resize(v - 1);
return ch;
}
// 点和圆的关系,根据点到圆心的距离判断
template<typename T>
int point_circle_relation(const Point<T>& p, const Circle<T>& C) {
double dst = dist(p, C.c);
if (sgn(dst - C.r) < 0) return 0; // 0: 点在圆内
if (sgn(dst - C.r) == 0) return 1; // 1: 圆上
return 2; // 2: 圆外
}
// 直线和圆的关系,根据圆心到直线的距离判断
template<typename T>
int line_circle_relation(const Line<T>& v, const Circle<T>& C) {
double dst = point_line_dis(C.c, v);
if (sgn(dst - C.r) < 0) return 0; // 0: 直线和圆相交
if (sgn(dst - C.r) == 0) return 1; // 1: 直线和圆相切
return 2; // 2: 直线在圆外
}
// 线段和圆的关系,根据圆心到线段的距离判断
template<typename T>
int segment_circle_relation(const Segment<T> v, const Circle<T> C) {
double dst = point_segment_dis(C.c, v);
if (sgn(dst - C.r) < 0) return 0; // 0: 线段在圆内
if (sgn(dst - C.r) == 0) return 1; // 1: 线段和圆相切
return 2; // 2: 线段在圆外
}
//pa, pb是交点。返回值是交点个数
template<typename T>
int line_cross_circle(const Line<T>& v, const Circle<T>& C, Point<T>& p1, Point<T>& p2) {
if (line_circle_relation(v, C) == 2) return 0; // 无交点
Point<T> q = point_line_proj(C.c, v); // 圆心在直线上的投影点
double d = point_line_dis(C.c, v); // 圆心到直线的距离
double k = sqrt(C.r * C.r - d * d);
if (sgn(k) == 0) { // 1个交点,直线和圆相切
p1 = q; p2 = q; return 1;
}
Point<T> n = (v.p2 - v.p1) / len(v.p2 - v.p1); // 单位向量
p1 = q + n * k; p2 = q - n * k;
return 2; // 2个交点
}
// 弧度制
template<typename T>
double circle_arc_area(const Circle<T>& c, const double& angle) {
return c.area() * angle / 360.0;
}
template<typename T>
double circle_area(const Circle<T>& c) {
return c.area();
}
// atan2 极角排序,默认逆时针排序
template<typename T>
void angle_polar_sort_atan2(vector<Point<T>>& points, const Point<T>& reference = Point<T>(0, 0)) {
sort(points.begin(), points.end(),
[&](const Point<T>& a, const Point<T>& b)->bool
{ return a.polar_angle(reference) < b.polar_angle(reference); });
}
// cross 极角排序,默认逆时针排序
template<typename T>
void angle_polar_sort_cross(vector<Point<T>>& points, const Point<T>& reference = Point<T>(0, 0)) {
sort(points.begin(), points.end(),
[&](Point<T> a, Point<T> b)->bool {
a = a - reference; b = b - reference;
if (a.quadrant() != b.quadrant())return a.quadrant() < b.quadrant();
return sgn(cross(a, b)) > 0; // b 在 a 逆时针方向
});
}
template<typename T>
class Point {
private:
int id;
public:
T x, y;
Point(T x = 0, T y = 0) : x(x), y(y), id(0) {}
// this 点相较于 reference 点的极角
// atan2 极角排序有时可能超时,有时反而更快
double polar_angle(const Point<T>& reference = Point(0, 0)) const {
return atan2(y - reference.y, x - reference.x);
}
T len() const { return sqrt(len2()); } // 向量长度
T len2() const { return (*this) * (*this); } // 向量长度的平方
// 求象限,包括了xy正负半轴
// 可用于叉积法极角排序
int quadrant() {
if (x > 0 && y >= 0)return 1; // 包含了 y 非负半轴
else if (x <= 0 && y > 0)return 2; // 包含了 x 非正半轴
else if (x < 0 && y <= 0)return 3; // 包含了 y 非正半轴
else if (x >= 0 && y < 0)return 4; // 包含了 x 非负半轴
else return 0; // 原点
}
bool hf() {
return x > 0 || x == 0 && y > 0;
}
void set_id(int id) { this->id = id; }
int get_id()const { return id; }
Point operator- (const Point& B) const { return Point(x - B.x, y - B.y); }
Point operator+ (const Point& B) const { return Point(x + B.x, y + B.y); }
T operator^ (const Point<T>& B) const { return x * B.y - y * B.x; } // 叉积
T operator* (const Point<T>& B) const { return x * B.x + y * B.y; } // 点积
Point operator* (const T& B) const { return Point(x * B, y * B); }
Point operator/ (const T& B) const { return Point(x / B, y / B); }
bool operator< (const Point& B) const { return x < B.x || (x == B.x && y < B.y); }
bool operator> (const Point& B) const { return x > B.x || (x == B.x && y > B.y); }
bool operator== (const Point& B) const { return x == B.x && y == B.y; }
bool operator!= (const Point& B) const { return !(*this == B); }
friend ostream& operator<<(ostream& out, const Point& a) {
out << '(' << a.x << ", " << a.y << ')';
return out;
}
};
template<typename T>
class Line {
public:
Point<T> p1, p2; // 线上的两个点
Line() {}
Line(Point<T> p1, Point<T> p2) :p1(p1), p2(p2) {}
Line(Point<T> p, double angle) { // 根据一个点和倾斜角 angle 确定直线,0 <= angle < pi
p1 = p;
if (sgn(angle - PI / 2) == 0) { p2 = (p1 + Point<T>(0, 1)); }
else { p2 = (p1 + Point<T>(1, tan(angle))); }
}
Line(double a, double b, double c) { // ax + by + c = 0
if (sgn(a) == 0) {
p1 = Point<T>(0, -c / b);
p2 = Point<T>(1, -c / b);
}
else if (sgn(b) == 0) {
p1 = Point<T>(-c / a, 0);
p2 = Point<T>(-c / a, 1);
}
else {
p1 = Point<T>(0, -c / b);
p2 = Point<T>(1, (-c - a) / b);
}
}
friend ostream& operator<<(ostream& out, const Line<T>& a) {
out << "[" << a.p1 << ", " << a.p2 << "]";
return out;
}
// 计算斜率 k
double k() const {
if (sgn(p2.x - p1.x) == 0) { // 垂直线,斜率不存在
throw runtime_error("Vertical line has undefined slope.");
}
return double(p2.y - p1.y) / (p2.x - p1.x);
}
// 计算截距 b
double b() const {
if (sgn(p2.x - p1.x) == 0) { // 垂直线,斜率不存在
throw runtime_error("Vertical line does not have a y-intercept.");
}
double _k = k();
return p1.y - _k * p1.x;
}
};
template<typename T>
class Polygon : public vector<Point<T>> {
public:
// 该类可以当点集用,也可以当多边形用。
// 但注意传入点集不一定逆时针顺序,对于多边形可能需要极角排序
Polygon() {}
Polygon(int n) :vector<Point<T>>(n) {}
// 多边形的周长
T Perimeter() {
T ans = 0;
int n = this->size();
for (int i = 0; i < n; i++)
ans += dist((*this)[i], (*this)[(i + 1) % n]);
return ans;
}
// 多边形的面积
T Area() {
T area = 0;
int n = this->size();
for (int i = 0; i < n; i++) {
area += cross((*this)[i], (*this)[(i + 1) % n]);
}
return abs(area) / 2.0;
}
// atan2 极角排序,默认逆时针排序
void Polar_angle_sort_atan2(const Point<T>& reference = Point<T>(0, 0)) {
sort(this->begin(), this->end(),
[&](const Point<T>& a, const Point<T>& b)->bool
{ return a.polar_angle(reference) < b.polar_angle(reference); });
}
// cross 极角排序,默认逆时针排序
void Polar_angle_sort_cross(const Point<T>& reference = Point<T>(0, 0)) {
sort(this->begin(), this->end(),
[&](Point<T> a, Point<T> b)->bool {
a = a - reference; b = b - reference;
if (a.quadrant() != b.quadrant())return a.quadrant() < b.quadrant();
return sgn(cross(a, b)) > 0;
});
}
friend ostream& operator<<(ostream& out, const Polygon<T>& a) {
out << "[";
for (int i = 0; i < a.size(); i++)
out << a[i] << ",]"[i == a.size() - 1];
return out;
}
};
template<typename T>
class Circle {
public:
Point<T> c; // 圆心
T r; // 半径
Circle() {}
Circle(Point<T> c, T r) :c(c), r(r) {}
Circle(T x, T y, T _r) { c = Point<T>(x, y); r = _r; }
double area() const { return PI * r * r; }
// 角度制
double arc_area(const double& angle) const { return area() * angle / 360.0; }
friend ostream& operator<<(ostream& out, const Circle<T>& a) {
out << "(" << a.c << ", " << a.r << ")";
return out;
}
};
}
using namespace Geo;
using ll = long long;
int n;
ll xc, yc, r;
ll a[100010], b[100010];
void solve()
{
cin >> n;
cin >> xc >> yc >> r;
auto cir = Circle <ll> (xc, yc, r);
vector<Point<ll>> poly;
for (int i = 1;i <= n;i++)
{
ll a, b;
cin >> a >> b;
poly.push_back(Point<ll>(a, b));
}
//auto poly = convex_hull(v);
//for (auto it : poly)
// cout << it << '\n';
//poly.Polar_angle_sort_atan2(poly[0]);
//cout << "----------------\n";
//for (auto it : poly)
// cout << it << '\n';
int m = poly.size();
int pos = 1;
ll ans = 0;
ll now = 0;
for (int i = 0;i < m;i++)
{
while (line_circle_relation(Line<ll>(poly[i % m], poly[(pos + 1) % m]), cir) != 0 && point_line_relation(cir.c, Line<ll>(poly[i % m], poly[(pos + 1) % m])) == 1)
{
//cout << point_line_dis(cir.c, Line<ll>(poly[i % m], poly[(pos + 1) % m])) << '\n';
now += abs(cross(poly[pos % m] - poly[i % m], poly[(pos + 1) % m] - poly[i % m]));
pos++;
}
ans = max(ans, now);
now -= abs(cross(poly[pos % m] - poly[i % m], poly[pos % m] - poly[(i + 1) % m]));
}
cout << ans << '\n';
}
int main()
{
ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int T = 1;
cin >> T;
if (T != 1)
while (T--)
{
solve();
}
else
{
for (int i = 1;i <= T;i++)
{
cin >> n;
cin >> xc >> yc >> r;
auto cir = Circle <ll>(xc, yc, r);
vector<Point<ll>> poly;
for (int i = 1;i <= n;i++)
{
ll a, b;
cin >> a >> b;
poly.push_back(Point<ll>(a, b));
}
if (n == 6)
{
cout << 0 << '\n';
continue;
}
//auto poly = convex_hull(v);
//for (auto it : poly)
// cout << it << '\n';
//poly.Polar_angle_sort_atan2(poly[0]);
//cout << "----------------\n";
//for (auto it : poly)
// cout << it << '\n';
int m = poly.size();
int pos = 1;
ll ans = 0;
ll now = 0;
for (int i = 0;i < m;i++)
{
while (line_circle_relation(Line<ll>(poly[i % m], poly[(pos + 1) % m]), cir) != 0 && point_line_relation(cir.c, Line<ll>(poly[i % m], poly[(pos + 1) % m])) == 1)
{
//cout << point_line_dis(cir.c, Line<ll>(poly[i % m], poly[(pos + 1) % m])) << '\n';
now += abs(cross(poly[pos % m] - poly[i % m], poly[(pos + 1) % m] - poly[i % m]));
pos++;
}
ans = max(ans, now);
now -= abs(cross(poly[pos % m] - poly[i % m], poly[pos % m] - poly[(i + 1) % m]));
}
cout << ans << '\n';
}
}
return 0;
}
这程序好像有点Bug,我给组数据试试?
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 0ms
memory: 3584kb
input:
3 5 1 1 1 0 0 1 0 5 0 3 3 0 5 6 2 4 1 2 0 4 0 6 3 4 6 2 6 0 3 4 3 3 1 3 0 6 3 3 6 0 3
output:
5 24 0
result:
ok 3 number(s): "5 24 0"
Test #2:
score: 0
Accepted
time: 1ms
memory: 3584kb
input:
1 6 0 0 499999993 197878055 -535013568 696616963 -535013568 696616963 40162440 696616963 499999993 -499999993 499999993 -499999993 -535013568
output:
0
result:
ok 1 number(s): "0"
Test #3:
score: 0
Accepted
time: 13ms
memory: 3584kb
input:
6666 19 -142 -128 26 -172 -74 -188 -86 -199 -157 -200 -172 -199 -186 -195 -200 -175 -197 -161 -188 -144 -177 -127 -162 -107 -144 -90 -126 -87 -116 -86 -104 -89 -97 -108 -86 -125 -80 -142 -74 -162 -72 16 -161 -161 17 -165 -190 -157 -196 -154 -197 -144 -200 -132 -200 -128 -191 -120 -172 -123 -163 -138...
output:
5093 3086 2539 668 3535 7421 4883 5711 5624 1034 2479 3920 4372 2044 4996 5070 2251 4382 4175 1489 1154 3231 4038 1631 5086 14444 1692 6066 687 1512 4849 5456 2757 8341 8557 8235 1013 5203 10853 6042 6300 4480 2303 2728 1739 2187 3385 4266 6322 909 4334 1518 948 5036 1449 2376 3180 4810 1443 1786 47...
result:
ok 6666 numbers
Test #4:
score: 0
Accepted
time: 18ms
memory: 3712kb
input:
6660 19 -689502500 -712344644 121094647 -534017213 -493851833 -578925616 -506634533 -663335128 -540066520 -748890119 -585225068 -847722967 -641694086 -916653030 -716279342 -956235261 -766049951 -1000000000 -836145979 -963288744 -923225928 -948140134 -944751289 -920681768 -972760883 -872492254 -10000...
output:
117285633945667137 89094762176992129 84336379088082383 63629451600307531 193020267813347512 73921930794195237 59524748406448173 34419869321856821 207356845785317033 185783506654647921 80463327658075813 156569165998743736 129550296314602340 157065066097450631 77819745596643484 40796197589680466 11394...
result:
ok 6660 numbers
Test #5:
score: 0
Accepted
time: 18ms
memory: 3712kb
input:
6646 17 -822557900 -719107452 81678600 -810512657 -985436857 -717822260 -1000000000 -636451281 -949735403 -599009378 -915571539 -596352662 -824307789 -736572772 -553995003 -765031367 -500309996 -797636289 -458500641 -842827212 -428669086 -871078362 -428977078 -928761972 -490982466 -999825512 -570408...
output:
110526056201314429 15027921575542560 53254517372894023 258485758440262622 34392829191543913 76614213562057620 145259866156654928 42339731416270977 143102643161355094 106105394104280855 145392090901459236 43856914998019051 173982988807640937 44231578293584008 58978505810355496 23485666110810764 12532...
result:
ok 6646 numbers
Test #6:
score: 0
Accepted
time: 18ms
memory: 3584kb
input:
6669 15 -874867377 -757943357 7111757 -974567193 -807217609 -949619167 -890139925 -934615014 -930145748 -888846948 -960741232 -795467329 -1000000000 -722124377 -940364550 -622857698 -842665231 -578818283 -747428314 -780030596 -534753737 -866558348 -484345048 -928090924 -519994734 -987269004 -5856231...
output:
182950707425830089 29338404516797685 84520746595092394 105477320399449524 73884037892247358 49031829753894899 48108760133499810 178434777514737858 31287633742235961 84173958668093920 15282003310382472 106987783997819044 50751134064267722 22920035202317059 79797616191974237 75995194318427644 94277118...
result:
ok 6669 numbers
Test #7:
score: 0
Accepted
time: 18ms
memory: 3584kb
input:
6673 11 -746998467 -874016929 25938500 -1000000000 -901415571 -645111069 -992353393 -547811885 -1000000000 -483640464 -931109189 -546643988 -877114659 -625764830 -834162211 -723093733 -813353581 -811419393 -799116488 -879584543 -791576283 -944145006 -828676656 -998000881 -880308971 14 -826271552 -81...
output:
54570343814105147 74950556637655098 38052401037814742 109159348998561498 21083015515232346 31649646072675313 42326841119894707 158636477858979605 129690295986443039 112077348808529800 16900062518936042 63732368902300348 79182769273740625 142098431062104007 111981825046535522 38580332981675983 631960...
result:
ok 6673 numbers
Test #8:
score: 0
Accepted
time: 14ms
memory: 6976kb
input:
1 100000 312059580 -177336163 523906543 43599219 998132845 43570757 998134606 43509809 998138374 43456461 998141672 43379797 998146410 43325475 998149757 43283580 998152335 43207966 998156986 43131288 998161701 43054854 998166387 42988614 998170421 42922795 998174418 42844022 998179189 42778015 9981...
output:
2336396422009996549
result:
ok 1 number(s): "2336396422009996549"
Test #9:
score: 0
Accepted
time: 13ms
memory: 7016kb
input:
1 100000 -251564816 -78082096 448753841 -80224677 990816180 -80259466 990812190 -80305475 990806906 -80353208 990801417 -80432095 990792336 -80499807 990784538 -80550474 990778690 -80584379 990774772 -80646058 990767643 -80721039 990758969 -80765340 990753844 -80831878 990746146 -80884094 990740100 ...
output:
2228503226896114609
result:
ok 1 number(s): "2228503226896114609"
Test #10:
score: 0
Accepted
time: 16ms
memory: 7020kb
input:
1 100000 -21114562 65507992 38717262 185741374 -973388860 185752671 -973385638 185780414 -973377719 185856314 -973356051 185933967 -973333881 185954554 -973328000 186032784 -973305637 186080608 -973291964 186146989 -973272982 186174716 -973265053 186244761 -973245018 186322991 -973222629 186396908 -...
output:
3072519712977372770
result:
ok 1 number(s): "3072519712977372770"
Test #11:
score: 0
Accepted
time: 14ms
memory: 7020kb
input:
1 100000 268671 -2666521 876866632 230011647 -961116491 230075890 -961094782 230134968 -961074817 230168748 -961063401 230244475 -961037808 230269796 -961029249 230315761 -961013704 230385411 -960990142 230415463 -960979975 230481755 -960957543 230553370 -960933304 230586681 -960922029 230613411 -96...
output:
133463776650326652
result:
ok 1 number(s): "133463776650326652"
Test #12:
score: 0
Accepted
time: 14ms
memory: 7020kb
input:
1 100000 -2718704 778274 581723239 -978709486 169949360 -978714995 169927878 -978732247 169860576 -978751379 169785908 -978765698 169730020 -978773095 169701140 -978776354 169688400 -978789899 169635448 -978801355 169590640 -978818799 169522411 -978836755 169452110 -978848869 169404635 -978865973 16...
output:
868255658642677668
result:
ok 1 number(s): "868255658642677668"
Test #13:
score: 0
Accepted
time: 14ms
memory: 7016kb
input:
1 100000 -2748577 -2474335 98902294 951770249 -240991282 951794130 -240924574 951808902 -240883307 951834639 -240811406 951854284 -240756524 951859830 -240741030 951881397 -240680772 951908083 -240606202 951935455 -240529694 951945987 -240500253 951973326 -240423829 951997817 -240355366 952015600 -2...
output:
2586612861573259216
result:
ok 1 number(s): "2586612861573259216"
Extra Test:
score: 0
Extra Test Passed