#include <bits/stdc++.h>
#define pf printf
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define sffff(a,b,c,d) scanf("%d %d %d %d",&a,&b,&c,&d)
#define sffffl(a,b,c,d) scanf("%lld %lld %lld %lld",&a,&b,&c,&d)
#define sfffff(a,b,c,d,e) scanf("%d %d %d %d %d",&a,&b,&c,&d,&e)
#define sfffffl(a,b,c,d,e) scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)
#define sfc(a) scanf("%c",&a)
#define ms(a,b) memset(a,b,sizeof(a))
#define pb(a) push_back(a)
#define pbp(a,b) push_back({a,b})
#define db double
#define ft float
#define ll long long
#define ull unsigned long long
#define ff first
#define ss second
#define sz(x) x.size()
#define qu queue
#define pqu priority_queue
#define vc vector
#define vi vector<int>
#define vll vector<long long>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define all(x) x.begin(),x.end()
#define CIN ios_base::sync_with_stdio(0); cin.tie(0)
#define max3(a, b, c) max(a, b) > max(b, c) ? max(a, b) : max(b, c)
#define min3(a, b, c) min(a, b) < min(b, c) ? min(a, b) : min(b, c)
#define loop0(i,n) for(int i=0;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define loopcmp(i,n) for(int i=1;i<n;i++)
#define looprev(i,n) for(int i=n-1; i>=0; i--)
#define loopab(i,a,b) for(int i=a;i<=b;i++)
#define loopba(i,b,a) for(int i=b;i>=a;i--)
#define stlloop(x) for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
#define case1(z) cout<<"Case "<<z<<": "
#define case2(z) printf("Case %d: ",z)
#define PI acos(-1) //3.14159265358979323846264338328
#define valid(tx,ty) tx>=0 && tx<row && ty>=0 && ty<col
#define intlim 2147483648
#define MAX 1000000
#define inf 100000000
/*------------------------------Graph Moves----------------------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move
//const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move
/*---------------------------------------------------------------------*/
using namespace std;
/*----------------------Matrix-----------------------*/
ll MOD=10007;
struct matrix
{
ll mat[4][4];
int row,col;
matrix()
{
ms(mat,0);
}
matrix(int a, int b)
{
row=a,col=b;
ms(mat,0);
}
matrix operator*(const matrix &p) const
{
assert(col == p.row);
matrix result;
result.row = row;
result.col = p.col;
for (int i = 0; i < result.row; i++)
{
for (int j = 0; j < result.col; j++)
{
ll sum = 0;
for (int k = 0; k <col; k++)
{
sum += ((mat[i][k]%MOD) * (p.mat[k][j]%MOD))%MOD;
sum%=MOD;
}
result.mat[i][j] = sum;
}
}
return result;
}
matrix operator+ (const matrix &p) const
{
assert(row==p.row && col==p.col);
matrix result;
result.row=row;
result.col=col;
for(int i=0; i<result.row; i++)
{
for(int j=0; j<result.col; j++)
result.mat[i][j]=((mat[i][j]%MOD)+(p.mat[i][j]%MOD))%MOD;;
}
return result;
}
matrix identity()
{
matrix result;
result.row=row;
result.col=col;
for(int i=0; i<row; i++)
result.mat[i][i]=1;
return result;
}
matrix pow(ll pow)
{
matrix result=(*this);
matrix ret=(*this).identity();
while(pow)
{
if(pow % 2==1)
ret=ret*result;
result=result*result;
pow/=2;
}
return ret;
}
void show()
{
printf("-----------------------------\n");
for(int i=0; i<row; i++)
{
for(int j=0; j<col; j++)
printf("%lld ",mat[i][j]);
printf("\n");
}
printf("-----------------------------\n");
}
};
ll power(int n,int p)
{
if(p==0) return 1;
ll r=1;
for(int i=0;i<p;i++) r*=n;
return r;
}
/*--------------------------Matrix End---------------------*/
int main()
{
///freopen("in.txt","r",stdin);
///freopen("out.txt","w",stdout);
ll n,a,b,c,t;
sfl(t);
loop1(z,t)
{
sffffl(n,a,b,c);
matrix base(4,4), multiplicand(4,1);
base.mat[0][0]=a;
base.mat[0][2]=b;
base.mat[0][3]=c;
base.mat[1][0]=base.mat[2][1]=base.mat[3][3]=1;
multiplicand.mat[3][0]=1;
base=base.pow(n-2);
base=base*multiplicand;
ll rs=base.mat[0][0];
if(n<=2) rs=0;
case1(z);
cout<<rs<<endl;
}
return 0;
}
Category Light OJ
Light OJ – 1065 – Number Sequence
In this problem, the first confusing thing is that, where is the MOD value !! But if you carefully read the problem statement, the MOD value is there indirectly as input ! 😀
When we have to find the last 1 digit of a given integer n, then the last digit k is-
k=n%10
or, k=n%10^1
Similarly when we are to find the last 1 digit of a given integer n, then –
k=n%100
or, k=n%10^2
similarly for finding the last m digit of a given integer n, then –
or, k=n%10^m
And here, this 10^m is the MOD value !! 😀
#include <bits/stdc++.h>
#define pf printf
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define sffff(a,b,c,d) scanf("%d %d %d %d",&a,&b,&c,&d)
#define sffffl(a,b,c,d) scanf("%lld %lld %lld %lld",&a,&b,&c,&d)
#define sfffff(a,b,c,d,e) scanf("%d %d %d %d %d",&a,&b,&c,&d,&e)
#define sfffffl(a,b,c,d,e) scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)
#define sfc(a) scanf("%c",&a)
#define ms(a,b) memset(a,b,sizeof(a))
#define pb(a) push_back(a)
#define pbp(a,b) push_back({a,b})
#define db double
#define ft float
#define ll long long
#define ull unsigned long long
#define ff first
#define ss second
#define sz(x) x.size()
#define qu queue
#define pqu priority_queue
#define vc vector
#define vi vector<int>
#define vll vector<long long>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define all(x) x.begin(),x.end()
#define CIN ios_base::sync_with_stdio(0); cin.tie(0)
#define max3(a, b, c) max(a, b) > max(b, c) ? max(a, b) : max(b, c)
#define min3(a, b, c) min(a, b) < min(b, c) ? min(a, b) : min(b, c)
#define loop0(i,n) for(int i=0;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define loopcmp(i,n) for(int i=1;i<n;i++)
#define looprev(i,n) for(int i=n-1; i>=0; i--)
#define loopab(i,a,b) for(int i=a;i<=b;i++)
#define loopba(i,b,a) for(int i=b;i>=a;i--)
#define stlloop(x) for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
#define case1(z) cout<<"Case "<<z<<": "
#define case2(z) printf("Case %d: ",z)
#define PI acos(-1) //3.14159265358979323846264338328
#define valid(tx,ty) tx>=0 && tx<row && ty>=0 && ty<col
#define intlim 2147483648
#define MAX 1000000
#define inf 100000000
/*------------------------------Graph Moves----------------------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move
//const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move
/*---------------------------------------------------------------------*/
using namespace std;
/*----------------------Matrix-----------------------*/
ll MOD;
struct matrix
{
ll mat[2][2];
int row,col;
matrix()
{
ms(mat,0);
}
matrix(int a, int b)
{
row=a,col=b;
ms(mat,0);
}
matrix operator*(const matrix &p) const
{
assert(col == p.row);
matrix result;
result.row = row;
result.col = p.col;
for (int i = 0; i < result.row; i++)
{
for (int j = 0; j < result.col; j++)
{
ll sum = 0;
for (int k = 0; k <col; k++)
{
sum += ((mat[i][k]%MOD) * (p.mat[k][j]%MOD))%MOD;
sum%=MOD;
}
result.mat[i][j] = sum;
}
}
return result;
}
matrix operator+ (const matrix &p) const
{
assert(row==p.row && col==p.col);
matrix result;
result.row=row;
result.col=col;
for(int i=0; i<result.row; i++)
{
for(int j=0; j<result.col; j++)
result.mat[i][j]=((mat[i][j]%MOD)+(p.mat[i][j]%MOD))%MOD;;
}
return result;
}
matrix identity()
{
matrix result;
result.row=row;
result.col=col;
for(int i=0; i<row; i++)
result.mat[i][i]=1;
return result;
}
matrix pow(ll pow)
{
matrix result=(*this);
matrix ret=(*this).identity();
while(pow)
{
if(pow % 2==1)
ret=ret*result;
result=result*result;
pow/=2;
}
return ret;
}
void show()
{
printf("-----------------------------\n");
for(int i=0; i<row; i++)
{
for(int j=0; j<col; j++)
printf("%lld ",mat[i][j]);
printf("\n");
}
printf("-----------------------------\n");
}
};
ll power(int n,int p)
{
if(p==0) return 1;
ll r=1;
for(int i=0;i<p;i++) r*=n;
return r;
}
/*--------------------------Matrix End---------------------*/
int main()
{
///freopen("in.txt","r",stdin);
///freopen("out.txt","w",stdout);
ll a,b,n,m,t;
sfl(t);
loop1(z,t)
{
sffffl(a,b,n,m);
MOD=power(10,m);
matrix base(2,2), multiplicand(2,1);
base.mat[0][0]=base.mat[0][1]=base.mat[1][0]=1;
multiplicand.mat[0][0]=b;
multiplicand.mat[1][0]=a;
base=base.pow(n-1);
base=base*multiplicand;
ll rs=base.mat[0][0];
if(n==0) rs=a%MOD;
else if(n==1) rs=b%MOD;
case1(z);
cout<<rs<<endl;
}
return 0;
}
Light OJ – 1305 – Area of a Parallelogram
This problem is too straight cut problem, no complexity, we have learned this within our Higher Secondary level 😀 It has to use those two formulas for solving two things in this problem-
1. Co-ordinates of the fourth point of a parallelogram,when its 3 co-ordinates are
known, is-
Dx=(Ax+Cx)-Bx …….(1)
Dy=(Ay+Cy)-By…….(2)
2. Area of a parallelogram when its 4 co-ordinates are known, is-
Area =0.5*( (Ax*By)-(Ay*Bx)+(Bx*Cy)-(By*Cx)+(Cx*Dy)-(Cy*Dx)+(Dx*Ay)-(Dy*Ax))…….(3)
Now just simply use these 3 formulas 😀
#include <bits/stdc++.h>
#define pf printf
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define sffff(a,b,c,d) scanf("%d %d %d %d",&a,&b,&c,&d)
#define sffffl(a,b,c,d) scanf("%lld %lld %lld %lld",&a,&b,&c,&d)
#define sfffff(a,b,c,d,e) scanf("%d %d %d %d %d",&a,&b,&c,&d,&e)
#define sfffffl(a,b,c,d,e) scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)
#define sfc(a) scanf("%c",&a)
#define ms(a,b) memset(a,b,sizeof(a))
#define pb(a) push_back(a)
#define pbp(a,b) push_back({a,b})
#define db double
#define ft float
#define ll long long
#define ull unsigned long long
#define ff first
#define ss second
#define sz(x) x.size()
#define qu queue
#define pqu priority_queue
#define vc vector
#define vi vector<int>
#define vll vector<long long>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define all(x) x.begin(),x.end()
#define CIN ios_base::sync_with_stdio(0); cin.tie(0)
#define max3(a, b, c) max(a, b) > max(b, c) ? max(a, b) : max(b, c)
#define min3(a, b, c) min(a, b) < min(b, c) ? min(a, b) : min(b, c)
#define loop0(i,n) for(int i=0;i<n;i++)
#define loopn(i,n) for(int i=1;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define loopi(i,n) for(int i=0;i<n-1;i++)
#define loopab(i,a,b) for(int i=a;i<=b;i++)
#define loopba(i,b,a) for(int i=b;i>=a;i--)
#define REV(i,n) for(i=n; i>=0; i--)
#define stlloop(x) for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
#define case1(z) cout<<"Case "<<z<<": "
#define case2(z) printf("Case %d: ",z)
#define PI acos(-1) //3.14159265358979323846264338328
#define valid(tx,ty) tx>=0 && tx<row && ty>=0 && ty<col
#define intlim 2147483648
#define MAX 1000000
#define inf 100000000
/*------------------------------Graph Moves----------------------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move
//const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move
/*---------------------------------------------------------------------*/
using namespace std;
int main()
{
//CIN;
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int t,ax,ay,bx,by,cx,cy,dx,dy,area;
sf(t);
loop1(z,t)
{
sfffff(ax,ay,bx,by,cx);
sf(cy);
dx=(ax+cx)-bx;
dy=(ay+cy)-by;
area=0.5*((ax*by)-(ay*bx)+(bx*cy)-(by*cx)+(cx*dy)-(cy*dx)+(dx*ay)-(dy*ax));
case2(z);
pf("%d %d %d\n",dx,dy,abs(area));
}
return 0;
}
Light OJ – 1072 – Calm Down
HINT :
This problem needs just simple trigonometric ratio. Think about this and if you can’t solve yet, then read further for the solution.
SOLUTION IDEA :
In the first figure, the circle contains 6 small circles . So the angle is (PI/6) . When the circle contains n small circles then the angle will be (PI/n) .
Now from trigonometry , we can write,
sin(PI/n) = r/(R-r)……………(1)
Now just by solving the equation (1) we get,
r = sin(PI/n)*(R-r)
r =( R*sin(PI/n)) – (r*sin(PI/n)
1 =(( R*sin(PI/n))/r) – sin(PI/n) [ dividing both sides by r ]
1 + sin(PI/n) = ( R*sin(PI/n))/r
r = ( R*sin(PI/n)) / (1 + sin(PI/n)) ……..(2)
Now, we have all the values of right side of equation (2). So just drop the values in (2) and get r !! 😀
#include <bits/stdc++.h>
#define pf printf
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define sffff(a,b,c,d) scanf("%d %d %d %d",&a,&b,&c,&d)
#define sffffl(a,b,c,d) scanf("%lld %lld %lld %lld",&a,&b,&c,&d)
#define sfffff(a,b,c,d,e) scanf("%d %d %d %d %d",&a,&b,&c,&d,&e)
#define sfffffl(a,b,c,d,e) scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)
#define ms(a,b) memset(a,b,sizeof(a))
#define pb(a) push_back(a)
#define db double
#define ft float
#define ll long long
#define ull unsigned long long
#define ff first
#define ss second
#define sz(x) x.size()
#define qu queue
#define pqu priority_queue
#define vc vector
#define vi vector<int>
#define vll vector<long long>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define all(x) x.begin(),x.end()
#define CIN ios_base::sync_with_stdio(0); cin.tie(0)
#define loop0(i,n) for(int i=0;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define loopab(a,b) for(int i=a;i<=b;i++)
#define stlloop(x) for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
#define case1(z) cout<<"Case "<<z<<": "
#define case2(z,x) cout<<"Case "<<z<<": "<<x<<endl
#define PI acos(-1) //3.14159265358979323846264338328
#define valid(tx,ty) tx>=0 && tx<r && ty>=0 && ty<c
#define intlim 2147483648
#define MAX 2000
#define inf 10000000
/*------------------------------Graph Moves----------------------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move
//const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move
/*---------------------------------------------------------------------*/
using namespace std;
int main()
{
//CIN;
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int t;
sf(t);
loop1(z,t)
{
db R;
int n;
scanf("%lf %d",&R,&n);
db angle=sin(PI/n);
db r=(angle*R)/(1+angle);
pf("Case %d: %.6f\n",z,r);
}
return 0;
}
Light OJ – 1043 – Triangle Partitioning
This problem is simply the theorem 3.10 of Higher Geometry – class 9-10 !! 😀
Let ADE / BDEC = x/y .
So ADE / ABC = x/(x+y)
From the theorem,we know that,
ADE / ABC = DE^2 / BC^2
or, sqrt (ADE / ABC) = DE/BC
or, sqrt (x/(x+y)) = DE/BC
or, r = DE/BC where r = sqrt(m)/sqrt((m+n))
As ADE and ABC are similar triangle, so
AD/AB = DE /BC
or, AD = DE/BC* AB
or, AD = r * AB .
Thus, the value of AD can be found simply by using the above formula.
NOTE: This problem also can be solved by “bisection” or other method if you don’t wanna bother remembering the formula.
#include <bits/stdc++.h>
#define pf printf
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define sffff(a,b,c,d) scanf("%d %d %d %d",&a,&b,&c,&d)
#define sffffl(a,b,c,d) scanf("%lld %lld %lld %lld",&a,&b,&c,&d)
#define sfffff(a,b,c,d,e) scanf("%d %d %d %d %d",&a,&b,&c,&d,&e)
#define sfffffl(a,b,c,d,e) scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)
#define sfc(a) scanf("%c",&a)
#define ms(a,b) memset(a,b,sizeof(a))
#define pb(a) push_back(a)
#define pbp(a,b) push_back({a,b})
#define db double
#define ft float
#define ll long long
#define ull unsigned long long
#define ff first
#define ss second
#define sz(x) x.size()
#define qu queue
#define pqu priority_queue
#define vc vector
#define vi vector<int>
#define vll vector<long long>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define all(x) x.begin(),x.end()
#define CIN ios_base::sync_with_stdio(0); cin.tie(0)
#define max3(a, b, c) max(a, b) > max(b, c) ? max(a, b) : max(b, c)
#define min3(a, b, c) min(a, b) < min(b, c) ? min(a, b) : min(b, c)
#define loop0(i,n) for(int i=0;i<n;i++)
#define loopn(i,n) for(int i=1;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define loopi(i,n) for(int i=0;i<n-1;i++)
#define loopab(i,a,b) for(int i=a;i<=b;i++)
#define loopba(i,b,a) for(int i=b;i>=a;i--)
#define REV(i,n) for(i=n; i>=0; i--)
#define stlloop(x) for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
#define case1(z) cout<<"Case "<<z<<": "
#define case2(z) printf("Case %d: ",z)
#define PI 3.14159265358979323846264338328
#define valid(tx,ty) tx>=0 && tx<row && ty>=0 && ty<col
#define intlim 2147483648
#define MAX 1000000
#define inf 100000000
/*------------------------------Graph Moves----------------------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move
//const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move
/*---------------------------------------------------------------------*/
using namespace std;
int main()
{
//CIN;
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int t;
sf(t);
loop1(z,t)
{
double AB,AC,BC,rat,x,AD;
cin>>AB>>AC>>BC>>rat;
x=sqrt((rat)/(rat+1)); // ADE/ABC=DE^2/BC^2 AND Here, ADE/ABC=(rat)/(rat+1) and x^2=DE^2/BC^2
AD=x*AB;
case2(z);
pf("%.6f\n",AD);
}
return 0;
}
Fall out, Codes! 