Hi guys,↵
↵
↵
I gave the Newton's coding challenge held today and could solve only 4, I got 15+ WAs on 5th and couldn't solve the 6th problem.↵
Can anyone please tell me the solution for the last problem and if possible tell me the error in my code for 5th problem?↵
↵
↵
Also feel free to use the comment section to discuss other solutions as well.↵
↵
[Please visit this link to see a screenshot of the problem statement of P5](https://ibb.co/HFDnLBR)↵
↵
[Please visit this link to see a screenshot of the problem statement of P6](https://ibb.co/MsNzpS1)↵
↵
<spoiler summary="My Code For P5">↵
~~~~~↵
//Nightmare05↵
#include<bits/stdc++.h>↵
↵
using namespace std;↵
↵
#define sp << " " <<↵
#define mod 1000000007↵
#define mp make_pair↵
#define pb push_back↵
#define int long long↵
#define double long double↵
#define INF (1e18+13)↵
#define PI 3.1415926535898↵
↵
int power(int p,int j)↵
{↵
int res=1;↵
p=p%mod;↵
while(j>0)↵
{↵
if(j&1)↵
res=(res*p)%mod;↵
j=j>>1;↵
p=(p*p)%mod;↵
}↵
return res;↵
}↵
↵
↵
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());↵
↵
int dice(int d,int p)↵
{↵
uniform_int_distribution<int> uid(d,p);//specify l and s.↵
return uid(rng) ;↵
}↵
↵
double s(double m1,double m2,double c1,double c2,double c3,double c4)↵
{↵
double x1,x2,x3,x4,y1,y2,y3,y4;↵
x1=(c2-c1)/(m1-m2);↵
x2=(c3-c2)/(m2-m1);↵
x3=(c4-c3)/(m1-m2);↵
x4=(c1-c4)/(m2-m1);↵
y1=m1*x1+c1;↵
y2=m2*x2+c2;↵
y3=m1*x3+c3;↵
y4=m2*x4+c4;↵
double d[4];↵
d[0]=(y2-y1)*(y2-y1)+(x2-x1)*(x2-x1);↵
d[1]=(y3-y2)*(y3-y2)+(x3-x2)*(x3-x2);↵
d[2]=(y4-y3)*(y4-y3)+(x4-x3)*(x4-x3);↵
d[3]=(y1-y4)*(y1-y4)+(x1-x4)*(x1-x4);↵
sort(d,d+4);↵
if(abs(d[3]-d[0])<0.01)↵
return sqrt((d[0]+d[3])/2)+26.0;↵
else↵
return -1;↵
}↵
↵
int32_t main()↵
{↵
ios::sync_with_stdio(false);↵
cin.tie(0);↵
cout.tie(0);↵
//freopen("elimination_validation_input.txt", "r", stdin);↵
//freopen("output.txt", "w", stdout);↵
int t;↵
cin >> t;↵
while(t--)↵
{↵
double dp=-1;↵
double x1,x2,x3,x4,y1,y2,y3,y4;↵
pair<double,double> a[4];↵
cin >> a[0].first >> a[0].second >> a[1].first >> a[1].second >> a[2].first >> a[2].second >> a[3].first >> a[3].second;↵
for(int j=0;j<6;j++)↵
{↵
if(j==1||j==3||j==5)↵
swap(a[2],a[3]);↵
if(j==2||j==4)↵
swap(a[1],a[3]);↵
x1=a[0].first;↵
y1=a[0].second;↵
x2=a[1].first;↵
y2=a[1].second;↵
x3=a[2].first;↵
y3=a[2].second;↵
x4=a[3].first;↵
y4=a[3].second;↵
for(double i=0.0;i<=1.0;i+=0.0001)↵
{↵
double m1=i*PI;↵
m1=tan(m1);↵
double c1=y1-m1*x1,c3=y3-m1*x3;↵
double m2=-1/m1;↵
if(m1==0.00000)↵
m2=tan(PI/2);↵
double c2=y2-m2*x2,c4=y4-m2*x4;↵
int h=1;↵
double p[3];↵
p[0]=y2-m1*x2-c1;↵
p[1]=y3-m1*x3-c1;↵
p[2]=y4-m1*x4-c1;↵
if(abs(p[0])>abs(p[1])||abs(p[2])>abs(p[1]))↵
h=0;↵
sort(p,p+3);↵
for(int z=0;z<3;z++)↵
if(abs(p[z])<0.0000000001)↵
h=0;↵
if(p[0]*p[2]<0)↵
h=0;↵
p[0]=y1-m2*x1-c2;↵
p[1]=y3-m2*x3-c2;↵
p[2]=y4-m2*x4-c2;↵
if(abs(p[0])>abs(p[2])||abs(p[1])>abs(p[2]))↵
h=0;↵
sort(p,p+3);↵
for(int z=0;z<3;z++)↵
if(abs(p[z])<0.0000000001)↵
h=0;↵
if(p[0]*p[2]<0)↵
h=0;↵
p[0]=y2-m1*x2-c3;↵
p[1]=y1-m1*x1-c3;↵
p[2]=y4-m1*x4-c3;↵
if(abs(p[0])>abs(p[1])||abs(p[2])>abs(p[1]))↵
h=0;↵
sort(p,p+3);↵
for(int z=0;z<3;z++)↵
if(abs(p[z])<0.0000000001)↵
h=0;↵
if(p[0]*p[2]<0)↵
h=0;↵
p[0]=y2-m2*x2-c4;↵
p[1]=y3-m2*x3-c4;↵
p[2]=y1-m2*x1-c4;↵
if(abs(p[1])>abs(p[0])||abs(p[2])>abs(p[0]))↵
h=0;↵
sort(p,p+3);↵
for(int z=0;z<3;z++)↵
if(abs(p[z])<0.0000000001)↵
h=0;↵
if(p[0]*p[2]<0)↵
h=0;↵
if(h)↵
dp=max(dp,s(m1,m2,c1,c2,c3,c4));↵
}↵
}↵
if(dp<26)↵
cout << -1 << endl;↵
else↵
cout << fixed << setprecision(2) << dp << endl;↵
}↵
return 0;↵
}↵
~~~~~↵
</spoiler>↵
↵
I would be really thankful if someone can point out, exactly what edge cases did I miss!↵
↵
Cheers!!!↵
↵
↵
I gave the Newton's coding challenge held today and could solve only 4, I got 15+ WAs on 5th and couldn't solve the 6th problem.↵
Can anyone please tell me the solution for the last problem and if possible tell me the error in my code for 5th problem?↵
↵
↵
Also feel free to use the comment section to discuss other solutions as well.↵
↵
[Please visit this link to see a screenshot of the problem statement of P5](https://ibb.co/HFDnLBR)↵
↵
[Please visit this link to see a screenshot of the problem statement of P6](https://ibb.co/MsNzpS1)↵
↵
<spoiler summary="My Code For P5">↵
~~~~~↵
//Nightmare05↵
#include<bits/stdc++.h>↵
↵
using namespace std;↵
↵
#define sp << " " <<↵
#define mod 1000000007↵
#define mp make_pair↵
#define pb push_back↵
#define int long long↵
#define double long double↵
#define INF (1e18+13)↵
#define PI 3.1415926535898↵
↵
int power(int p,int j)↵
{↵
int res=1;↵
p=p%mod;↵
while(j>0)↵
{↵
if(j&1)↵
res=(res*p)%mod;↵
j=j>>1;↵
p=(p*p)%mod;↵
}↵
return res;↵
}↵
↵
↵
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());↵
↵
int dice(int d,int p)↵
{↵
uniform_int_distribution<int> uid(d,p);//specify l and s.↵
return uid(rng) ;↵
}↵
↵
double s(double m1,double m2,double c1,double c2,double c3,double c4)↵
{↵
double x1,x2,x3,x4,y1,y2,y3,y4;↵
x1=(c2-c1)/(m1-m2);↵
x2=(c3-c2)/(m2-m1);↵
x3=(c4-c3)/(m1-m2);↵
x4=(c1-c4)/(m2-m1);↵
y1=m1*x1+c1;↵
y2=m2*x2+c2;↵
y3=m1*x3+c3;↵
y4=m2*x4+c4;↵
double d[4];↵
d[0]=(y2-y1)*(y2-y1)+(x2-x1)*(x2-x1);↵
d[1]=(y3-y2)*(y3-y2)+(x3-x2)*(x3-x2);↵
d[2]=(y4-y3)*(y4-y3)+(x4-x3)*(x4-x3);↵
d[3]=(y1-y4)*(y1-y4)+(x1-x4)*(x1-x4);↵
sort(d,d+4);↵
if(abs(d[3]-d[0])<0.01)↵
return sqrt((d[0]+d[3])/2)+26.0;↵
else↵
return -1;↵
}↵
↵
int32_t main()↵
{↵
ios::sync_with_stdio(false);↵
cin.tie(0);↵
cout.tie(0);↵
//freopen("elimination_validation_input.txt", "r", stdin);↵
//freopen("output.txt", "w", stdout);↵
int t;↵
cin >> t;↵
while(t--)↵
{↵
double dp=-1;↵
double x1,x2,x3,x4,y1,y2,y3,y4;↵
pair<double,double> a[4];↵
cin >> a[0].first >> a[0].second >> a[1].first >> a[1].second >> a[2].first >> a[2].second >> a[3].first >> a[3].second;↵
for(int j=0;j<6;j++)↵
{↵
if(j==1||j==3||j==5)↵
swap(a[2],a[3]);↵
if(j==2||j==4)↵
swap(a[1],a[3]);↵
x1=a[0].first;↵
y1=a[0].second;↵
x2=a[1].first;↵
y2=a[1].second;↵
x3=a[2].first;↵
y3=a[2].second;↵
x4=a[3].first;↵
y4=a[3].second;↵
for(double i=0.0;i<=1.0;i+=0.0001)↵
{↵
double m1=i*PI;↵
m1=tan(m1);↵
double c1=y1-m1*x1,c3=y3-m1*x3;↵
double m2=-1/m1;↵
if(m1==0.00000)↵
m2=tan(PI/2);↵
double c2=y2-m2*x2,c4=y4-m2*x4;↵
int h=1;↵
double p[3];↵
p[0]=y2-m1*x2-c1;↵
p[1]=y3-m1*x3-c1;↵
p[2]=y4-m1*x4-c1;↵
if(abs(p[0])>abs(p[1])||abs(p[2])>abs(p[1]))↵
h=0;↵
sort(p,p+3);↵
for(int z=0;z<3;z++)↵
if(abs(p[z])<0.0000000001)↵
h=0;↵
if(p[0]*p[2]<0)↵
h=0;↵
p[0]=y1-m2*x1-c2;↵
p[1]=y3-m2*x3-c2;↵
p[2]=y4-m2*x4-c2;↵
if(abs(p[0])>abs(p[2])||abs(p[1])>abs(p[2]))↵
h=0;↵
sort(p,p+3);↵
for(int z=0;z<3;z++)↵
if(abs(p[z])<0.0000000001)↵
h=0;↵
if(p[0]*p[2]<0)↵
h=0;↵
p[0]=y2-m1*x2-c3;↵
p[1]=y1-m1*x1-c3;↵
p[2]=y4-m1*x4-c3;↵
if(abs(p[0])>abs(p[1])||abs(p[2])>abs(p[1]))↵
h=0;↵
sort(p,p+3);↵
for(int z=0;z<3;z++)↵
if(abs(p[z])<0.0000000001)↵
h=0;↵
if(p[0]*p[2]<0)↵
h=0;↵
p[0]=y2-m2*x2-c4;↵
p[1]=y3-m2*x3-c4;↵
p[2]=y1-m2*x1-c4;↵
if(abs(p[1])>abs(p[0])||abs(p[2])>abs(p[0]))↵
h=0;↵
sort(p,p+3);↵
for(int z=0;z<3;z++)↵
if(abs(p[z])<0.0000000001)↵
h=0;↵
if(p[0]*p[2]<0)↵
h=0;↵
if(h)↵
dp=max(dp,s(m1,m2,c1,c2,c3,c4));↵
}↵
}↵
if(dp<26)↵
cout << -1 << endl;↵
else↵
cout << fixed << setprecision(2) << dp << endl;↵
}↵
return 0;↵
}↵
~~~~~↵
</spoiler>↵
↵
I would be really thankful if someone can point out, exactly what edge cases did I miss!↵
↵
Cheers!!!↵
