QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#720034 | #9434. Italian Cuisine | OldMemory | WA | 48ms | 5840kb | C++20 | 20.3kb | 2024-11-07 10:21:49 | 2024-11-07 10:21:49 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
#define int long long
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----------*/
const int N = 1e5 + 10;
int bgpn;
Point bgp[N];
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() {
cin >> bgpn;
int xc, yc, r;
cin >> xc >> yc >> r;
cir = Circle(Point(xc, yc), r);
for(int i = 0; i < bgpn; i++) {
int x, y;
cin >> x >> y;
bgp[i] = Point(x, y);
}
if(bgpn <= 3) {
cout << 0 << '\n';
return;
}
chn = Convex_hull(bgp, bgpn, ch);
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]);
}
cout << ans << '\n';
}
signed main() {
ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int T = 1;
if(multi) cin >> T;
while(T--) {
solve();
}
return 0;
}
Details
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Test #1:
score: 100
Accepted
time: 1ms
memory: 5840kb
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: 5712kb
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: -100
Wrong Answer
time: 48ms
memory: 5716kb
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 4569 2251 4382 4175 1489 1154 3231 4038 1631 5086 14444 1692 6066 687 1512 4849 5456 2675 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 46...
result:
wrong answer 16th numbers differ - expected: '5070', found: '4569'