QOJ.ac
QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#720067 | #9434. Italian Cuisine | OldMemory | AC ✓ | 50ms | 6972kb | C++20 | 20.7kb | 2024-11-07 10:30:37 | 2024-11-07 10:30:37 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
#define int __int128_t
typedef pair<int, int> PII;
typedef long long LL;
const int INF = 0x3f3f3f3f3f3f3f3f;
#define double long double
bool multi = 1;
/*----------计算几何模板----------*/
const double eps = 1e-8,pi = acos(-1.0),inf = 1e30;
int sgn(double x){ //判断x的大小
if(fabs(x) < eps) return 0; //x==0,返回0
else return x<0?-1:1; //x<0返回-1,x>0返回1
}
int dcmp(double x, double y){ //比较两个浮点数
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
}
struct Point{
double x, y;
Point(){}
Point(double x,double y):x(x),y(y){}
Point operator+(Point B){return Point(x+B.x,y+B.y);}
Point operator-(Point B){return Point(x-B.x,y-B.y);}
Point operator*(double k){return Point(x*k,y*k);}
Point operator/(double k){return Point(x/k,y/k);}
bool operator==(Point B){return sgn(x-B.x)==0&&sgn(y-B.y)==0;}
bool operator < (Point B){ //用于sort()排序,先按x排序,再按y排序
return sgn(x-B.x)<0 || (sgn(x-B.x)==0 && sgn(y-B.y)<0);
}
};
typedef Point Vector;
double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}//向量点积
double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}//向量叉积
double Distance(Point A, Point B){ return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));}//两点间距离
double Distance2(Point A, Point B){ return (A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y);}//两点间距离
double Len(Vector A){return sqrt(Dot(A,A));}//向量长度
double Len2(Vector A){return Dot(A,A);}//向量长度的平方
double Angle(Vector A,Vector B){return acos(Dot(A,B)/Len(A)/Len(B));}//向量A与B的夹角
double Area2(Point A,Point B,Point C){return Cross(B-A,C-A);}//以A,B,C三点构成的平行四边形的面积
Vector Rotate(Vector A, double rad){//将向量A逆时针旋转角度rad
return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
}
Vector Normal(Vector A){return Vector(-A.y/Len(A), A.x/Len(A));}//求向量A的单位法向量(逆时针)
bool Parallel(Vector A, Vector B){return sgn(Cross(A,B)) == 0;}//判断两个向量是否平行或重合
struct Line{
Point p1,p2; //(1)线上的两个点
Line(){}
Line(Point p1,Point p2):p1(p1),p2(p2){}
Line(Point p,double angle){ //(4)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(0,1));}
else{p2 = (p1 + Point(1,tan(angle)));}
}
Line(double a,double b,double c){ //(2)ax+by+c=0
if(sgn(a) == 0){
p1 = Point(0,-c/b);
p2 = Point(1,-c/b);
}
else if(sgn(b) == 0){
p1 = Point(-c/a,0);
p2 = Point(-c/a,1);
}
else{
p1 = Point(0,-c/b);
p2 = Point(1,(-c-a)/b);
}
}
};
typedef Line Segment;
//点和直线的位置关系
int Point_line_relation(Point p, Line 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:0 点不在线段v上;1 点在线段v上
bool Point_on_seg(Point p, Line v){
return sgn(Cross(p-v.p1, v.p2-v.p1)) == 0 && sgn(Dot(p-v.p1,p-v.p2)) <= 0;
}
//求点到直线的距离
double Dis_point_line(Point p,Line v){
return fabs(Cross(p-v.p1,v.p2-v.p1))/Distance(v.p1,v.p2);
}
//点在直线上的投影
Point Point_line_proj(Point p, Line 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
Point Point_line_symmetry(Point p, Line v){
Point q = Point_line_proj(p,v);
return Point(2*q.x-p.x,2*q.y-p.y);
}
//点到线段的距离
double Dis_point_seg(Point p, Segment v){
if(sgn(Dot(p- v.p1,v.p2-v.p1))<0 || sgn(Dot(p- v.p2,v.p1-v.p2))<0)
return min(Distance(p,v.p1),Distance(p,v.p2));
return Dis_point_line(p,v); //点的投影在线段上
}
//两条直线的位置关系
int Line_relation(Line v1, Line 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
}
//两条直线的交点
Point Cross_point(Point a,Point b,Point c,Point 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(c.x*s2-d.x*s1,c.y*s2-d.y*s1)/(s2-s1);
}
//两个线段是否相交
bool Cross_segment(Point a,Point b,Point c,Point 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
int Point_in_polygon(Point pt,Point *p,int n){ //点pt,多边形Point *p
for(int i = 0;i < n;i++){ //3:点在多边形的顶点上
if(p[i] == pt) return 3;
}
for(int i = 0;i < n;i++){ //2:点在多边形的边上
Line v=Line(p[i],p[(i+1)%n]);
if(Point_on_seg(pt,v)) return 2;
}
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
}
//多边形的面积
double Polygon_area(Point *p, int n){ //Point *p表示多边形
double area = 0;
for(int i = 0;i < n;i++)
area += Cross(p[i],p[(i+1)%n]);
return area/2; //面积有正负,返回时不能简单地取绝对值
}
Point Polygon_center(Point *p, int n){ //求多边形重心
Point ans(0,0);
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;
}
//Convex_hull()求凸包。凸包顶点放在ch中,返回值是凸包的顶点数
int Convex_hull(Point *p,int n,Point *ch){
n = unique(p,p+n)-p; //去除重复点
sort(p,p+n); //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) //把后面ch[v-1]改成ch[v-2]也行
v--;
ch[v++]=p[i];
}
int j=v;
//求上凸包
for(int i=n-2;i>=0;i--){
while(v>j && sgn(Cross(ch[v-1]-ch[v-2],p[i]-ch[v-1]))<=0) //把后面ch[v-1]改成ch[v-2]也行
v--;
ch[v++]=p[i];
}
if(n>1) v--;
return v; //返回值v是凸包的顶点数
}
//返回p[left]~p[right]的平面最近点对距离
double Get_Closest_Pair(Point p[],int left,int right){
sort(p+left,p+right+1,[](Point A,Point B){
return sgn(A.x-B.x)<0 || (sgn(A.x-B.x)==0 && sgn(A.y-B.y)<0);
});
vector<Point> tmp_p(right-left+10);
function<double(int,int)> Closest_Pair=[&](int left,int right){
double dis = inf;
if(left == right) return dis; //只剩1个点
if(left + 1 == right) return Distance(p[left], p[right]);//只剩2个点
int mid = (left+right)/2; //分治
double d1 = Closest_Pair(left,mid); //求s1内的最近点对
double d2 = Closest_Pair(mid+1,right); //求s2内的最近点对
dis = min(d1,d2);
int k = 0;
for(int i=left;i<=right;i++) //在s1和s2中间附近找可能的最小点对
if(fabs(p[mid].x - p[i].x) <= dis) //按x坐标来找
tmp_p[k++] = p[i];
sort(&tmp_p[0],&tmp_p[k],[](Point A,Point B){return sgn(A.y-B.y)<0;}); //按y坐标排序,用于剪枝。这里不能按x坐标排序
for(int i=0;i<k;i++)
for(int j=i+1;j<k;j++){
if(tmp_p[j].y - tmp_p[i].y >= dis) break; //剪枝
dis = min(dis,Distance(tmp_p[i],tmp_p[j]));
}
return dis; //返回最小距离
};
return Closest_Pair(left,right);
}
double Rotating_Calipers2(Point p[], int left, int right) {
int n = right - left + 1;
if(n <= 2) {
return Distance2(p[left], p[right]);
}
double res = 0;
for(int i = 0, j = 2; i < n; i++) {
while(Area2(p[left + i], p[left + (i + 1) % n], p[left + j]) < Area2(p[left + i], p[left + (i + 1) % n], p[left + (j + 1) % n])) {
j = (j + 1) % n;
}
res = max({res, Distance2(p[left + i], p[left + j]), Distance2(p[left + (i + 1) % n], p[left + j])});
}
return res;
}
struct HLine{
Point p; //直线上一个点
Vector v; //方向向量,它的左边是半平面
double ang; //极角,从x正半轴旋转到v的角度
HLine(){};
HLine(Point p, Vector v):p(p),v(v){ang = atan2(v.y, v.x);}
bool operator < (HLine &L){return ang < L.ang;} //用于排序
};
//点p在线L左边,即点p在线L在外面:
bool OnLeft(HLine L,Point p){return sgn(Cross(L.v,p-L.p))>=0;}
Point Cross_point(HLine a,HLine b){ //两直线交点
Vector u=a.p-b.p;
double t=Cross(b.v,u)/Cross(a.v,b.v);
return a.p+a.v*t;
}
vector<Point> HPI(vector<HLine> L){ //求半平面交,返回凸多边形
int n=L.size();
sort(L.begin(),L.end()); //将所有半平面按照极角排序。
int first,last; //指向双端队列的第一个和最后一个元素
vector<Point> p(n); //两个相邻半平面的交点
vector<HLine> q(n); //双端队列
vector<Point> ans; //半平面交形成的凸包
q[first=last=0]=L[0];
for(int i=1;i<n;i++){
//情况1:删除尾部的半平面
while(first<last && !OnLeft(L[i], p[last-1])) last--;
//情况2:删除首部的半平面:
while(first<last && !OnLeft(L[i], p[first])) first++;
q[++last]=L[i]; //将当前的半平面加入双端队列尾部
//极角相同的两个半平面,保留左边:
if(fabs(Cross(q[last].v,q[last-1].v)) < eps){
last--;
if(OnLeft(q[last],L[i].p)) q[last]=L[i];
}
//计算队列尾部半平面交点:
if(first<last) p[last-1]=Cross_point(q[last-1],q[last]);
}
//情况3:删除队列尾部的无用半平面
while(first<last && !OnLeft(q[first],p[last-1])) last--;
if(last-first<=1) return ans; //空集
p[last]=Cross_point(q[last],q[first]); //计算队列首尾部的交点。
for(int i=first;i<=last;i++) ans.push_back(p[i]); //复制。
return ans; //返回凸多边形
}
struct Circle{
Point c; //圆心
double r; //半径
Circle(){}
Circle(Point c,double r):c(c),r(r){}
Circle(double x,double y,double _r){c=Point(x,y);r = _r;}
};
//点与圆的关系
int Point_circle_relation(Point p, Circle C){
double dst = Distance(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
}
#define i128 __int128_t
i128 MyDistance2(Point A, Point B){ return (A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y);}//两点间距离
//求点到直线的距离
i128 My_Dis_point_line(Point p,Line v, double d){
return (i128)round(fabs(Cross(p-v.p1,v.p2-v.p1))) * (i128)round(fabs(Cross(p-v.p1,v.p2-v.p1))) - (i128)round(d) * (i128)round(d) * MyDistance2(v.p1,v.p2);
}
//直线与圆的关系
int Line_circle_relation(Line v,Circle C){
i128 dst = My_Dis_point_line(C.c,v, C.r);
if(dst < 0) return 0; //0 直线和圆相交
if(dst ==0) return 1; //1 直线和圆相切
return 2; //2 直线在圆外
}
//线段与圆的关系
int Seg_circle_relation(Segment v,Circle C){
double dst = Dis_point_seg(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
int Line_cross_circle(Line v,Circle C,Point &pa,Point &pb){
if(Line_circle_relation(v, C)==2) return 0;//无交点
Point q = Point_line_proj(C.c,v); //圆心在直线上的投影点
double d = Dis_point_line(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
pa = q; pb = q; return 1;
}
Point n=(v.p2-v.p1)/ Len(v.p2-v.p1); //单位向量
pa = q + n*k; pb = q - n*k;
return 2; //2个交点
}
//三点确定圆心
Point circle_center(const Point a, const Point b, const Point c){
Point center;
double a1=b.x-a.x, b1=b.y-a.y, c1=(a1*a1+b1*b1)/2;
double a2=c.x-a.x, b2=c.y-a.y, c2=(a2*a2+b2*b2)/2;
double d =a1*b2-a2*b1;
center.x =a.x+(c1*b2-c2*b1)/d;
center.y =a.y+(a1*c2-a2*c1)/d;
return center;
}
//求最小圆覆盖
void min_cover_circle(Point *p, int n, Circle &C){
random_shuffle(p, p + n); //随机函数,打乱所有点。这一步很重要
Point c=p[0];
double r=0; //从第1个点p0开始。圆心为p0,半径为0
for(int i=1;i<n;i++) //扩展所有点
if(sgn(Distance(p[i],c)-r)>0){ //点pi在圆外部
c=p[i]; r=0; //重新设置圆心为pi,半径为0
for(int j=0;j<i;j++) //重新检查前面所有的点。
if(sgn(Distance(p[j],c)-r)>0){ //两点定圆
c.x=(p[i].x + p[j].x)/2;
c.y=(p[i].y + p[j].y)/2;
r=Distance(p[j],c);
for(int k=0;k<j;k++)
if (sgn(Distance(p[k],c)-r)>0){ //两点不能定圆,就三点定圆
c=circle_center(p[i],p[j],p[k]);
r=Distance(p[i], c);
}
}
}
C={c,r};
}
struct Point3{ //三维点
double x,y,z;
Point3(){}
Point3(double x,double y,double z):x(x),y(y),z(z){}
Point3 operator + (Point3 B){return Point3(x+B.x,y+B.y,z+B.z);}
Point3 operator - (Point3 B){return Point3(x-B.x,y-B.y,z-B.z);}
Point3 operator * (double k){return Point3(x*k,y*k,z*k);}
Point3 operator / (double k){return Point3(x/k,y/k,z/k);}
bool operator == (Point3 B){ return sgn(x-B.x)==0 && sgn(y-B.y)==0 && sgn(z-B.z)==0;}
};
typedef Point3 Vector3; //三维向量
double Distance(Vector3 A,Vector3 B){
return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)+ (A.z-B.z)*(A.z-B.z));
}
struct Line3{
Point3 p1,p2;
Line3(){}
Line3(Point3 p1,Point3 p2):p1(p1),p2(p2){}
};
typedef Line3 Segment3; //定义线段,两端点是Point p1,p2
//三维点积
double Dot(Vector3 A,Vector3 B){return A.x*B.x+A.y*B.y+A.z*B.z;}
//求向量A的长度
double Len(Vector3 A){ return sqrt(Dot(A, A));}
//求向量A的长度的平方
double Len2(Vector3 A){ return Dot(A, A);}
//求向量A与B的夹角
double Angle(Vector3 A,Vector3 B){return acos(Dot(A,B)/Len(A)/Len(B));}
//三维叉积
Vector3 Cross(Vector3 A,Vector3 B){
return Point3(A.y*B.z-A.z*B.y, A.z*B.x-A.x*B.z, A.x*B.y-A.y*B.x);
}
//求三个点构成的三角形面积的2倍
double Area2(Point3 A,Point3 B,Point3 C){return Len(Cross(B-A, C-A));}
//三维:点在直线上
bool Point_line_relation(Point3 p,Line3 v){
return sgn( Len(Cross(v.p1-p,v.p2-p))) == 0 && sgn(Dot(v.p1-p,v.p2-p))== 0;
}
double Dis_point_line(Point3 p, Line3 v) //三维:点到直线距离
{ return Len(Cross(v.p2-v.p1,p-v.p1))/Distance(v.p1,v.p2); }
//三维:点到线段距离。
double Dis_point_seg(Point3 p, Segment3 v){
if(sgn(Dot(p- v.p1,v.p2-v.p1)) < 0 || sgn(Dot(p- v.p2,v.p1-v.p2)) < 0)
return min(Distance(p,v.p1),Distance(p,v.p2));
return Dis_point_line(p,v);
}
//三维:点 p 在直线上的投影
Point3 Point_line_proj(Point3 p, Line3 v){
double k=Dot(v.p2-v.p1,p-v.p1)/Len2(v.p2-v.p1);
return v.p1+(v.p2-v.p1)*k;
}
struct Plane{
Point3 p1,p2,p3; //平面上的三个点
Plane(){}
Plane(Point3 p1,Point3 p2,Point3 p3):p1(p1),p2(p2),p3(p3){}
};
//平面的法向量
Point3 Pvec(Point3 A, Point3 B, Point3 C){return Cross(B-A,C-A);}
Point3 Pvec(Plane f){return Cross(f.p2-f.p1,f.p3-f.p1);}
//四点共平面
bool Point_on_plane(Point3 A,Point3 B,Point3 C,Point3 D){
return sgn(Dot(Pvec(A,B,C),D-A)) == 0;
}
//两平面平行
int Parallel(Plane f1, Plane f2){ return Len(Cross(Pvec(f1),Pvec(f2))) < eps; }
//两平面垂直
int Vertical(Plane f1, Plane f2){ return sgn(Dot(Pvec(f1),Pvec(f2)))==0; }
//直线与平面的交点
int Line_cross_plane(Line3 u,Plane f,Point3 &p){
Point3 v = Pvec(f); //平面的法向量
double x = Dot(v, u.p2-f.p1);
double y = Dot(v, u.p1-f.p1);
double d = x-y;
if(sgn(x) == 0 && sgn(y) == 0) return -1; //-1:v在f上
if(sgn(d) == 0) return 0; //0:v与f平行
p = ((u.p1 * x)-(u.p2 * y))/d; //v与f相交
return 1;
}
//四面体有向体积*6
double volume4(Point3 a,Point3 b,Point3 c,Point3 d){
return Dot(Cross(b-a,c-a),d-a); }
//最小球覆盖
double min_cover_ball(Point3 *p, int n){
double T=100.0; //初始温度
double delta = 0.98; //降温系数
Point3 c = p[0]; //球心
int pos;
double r; //半径
while(T>eps) { //eps是终止温度
pos = 0; r = 0; //初始:p[0]是球心,半径是0
for(int i=0; i<n; i++) //迭代:找距离球心最远的点
if(Distance(c,p[i])>r){
r = Distance(c,p[i]); //距离球心最远的点肯定在球周上
pos = i;
}
c.x += (p[pos].x-c.x)/r*T; //逼近最后的解
c.y += (p[pos].y-c.y)/r*T;
c.z += (p[pos].z-c.z)/r*T;
T *= delta; //降温
}
return r;
}
/*----------计算几何模板----------*/
template <typename T>
inline void read(T &x)
{
T f = 1;
x = 0;
char ch = getchar();
while (0 == isdigit(ch))
{
if (ch == '-')
f = -1;
ch = getchar();
}
while (0 != isdigit(ch))
x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
x *= f;
}
template <typename T>
inline void write(T x)
{
if (x < 0)
{
x = ~(x - 1);
putchar('-');
}
if (x > 9)
write(x / 10);
putchar(x % 10 + '0');
}
const int N = 1e5 + 10;
int chn;
Point ch[N];
Circle cir;
int myCross(Vector A,Vector B){return (int)round(A.x)*(int)round(B.y)-(int)round(A.y)*(int)round(B.x);}//向量叉积
void solve() {
read(chn);
int xc, yc, r;
read(xc), read(yc), read(r);
cir = Circle(Point(xc, yc), r);
for(int i = 0; i < chn; i++) {
int x, y;
read(x), read(y);
ch[i] = Point(x, y);
}
if(chn <= 3) {
cout << 0 << '\n';
return;
}
int ans = 0;
int curS = 0;
for(int i = 0, j = i; i < chn; i++) {
while(1) {
ans = max(ans, curS);
Point curp = ch[j], nep = ch[(j + 1) % chn];
Point cirp = cir.c;
HLine nel = HLine(cirp, ch[i] - cirp);
if(!OnLeft(nel, nep) || Line_circle_relation(Line(ch[i], nep), cir) == 0) {
break;
}
curS += Cross(curp, nep) + Cross(nep, ch[i]) - Cross(curp, ch[i]);
j = (j + 1) % chn;
}
// if(i == 0) {
// cout << ch[i].x << ' ' << ch[i].y << '\n';
// cout << ch[j].x << ' ' << ch[j].y << '\n';
// cout << curS << '\n';
// }
ans = max(ans, curS);
int nei = (i + 1) % chn;
curS += Cross(ch[j], ch[nei]) - Cross(ch[i], ch[nei]) - Cross(ch[j], ch[i]);
}
write(ans);
putchar('\n');
}
signed main() {
ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int T = 1;
if(multi) read(T);
while(T--) {
solve();
}
return 0;
}
这程序好像有点Bug,我给组数据试试?
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 3508kb
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: 0ms
memory: 3480kb
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: 38ms
memory: 3796kb
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: 48ms
memory: 3716kb
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: 44ms
memory: 3648kb
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: 48ms
memory: 3712kb
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: 48ms
memory: 3636kb
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: 44ms
memory: 6772kb
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: 43ms
memory: 6736kb
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: 49ms
memory: 6920kb
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: 44ms
memory: 6760kb
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: 43ms
memory: 6972kb
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: 50ms
memory: 6752kb
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