#include <stdio.h>
#include <stdlib.h>
#define PI 3.1415
int main()
{
int i,T;
scanf("%d",&T);
for(i=0;i<T;i++)
{
int L;
double b,r,x,y;
scanf("%d",&L);
b=(6*L*1.0)/(10*1.0);
r=(L*1.0)/(5*1.0);
x=(acos(-1)*r*r*1.0);
y=(L*b*1.0)-(x*1.0);
printf("%.2f %.2f\n",x,y);
}
return 0;
}
Category Geometry
UVa – 10347 – Medians
This problem is based on basic geometry. We normally deals with given three sides to find out the area of a valid triangle. Here, we are given three medians to find out the area. Now, it is almost same as for given 3 sides.
For a valid traingle,
1. given three sides – a, b & c :
Perimeter, s = (a+b+c)/2
Area = sqrt(s*(s-a)*(s-b)*(s-c) )
2. given three medians – a, b & c :
Perimeter, s = (a+b+c)/2
Area = (4/3)*( sqrt(s*(s-a)*(s-b)*(s-c) ) )
Now, for both 1 and 2 , condition for validity of the triangle is –
a+b>c && b+c>a && c+a>b
#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 pii pair<int,int>
#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 pii pair<int,int>
#define ff first
#define ss second
#define sz(x) x.size()
#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 for0(i,n) for(int i=0;i<n;i++)
#define for1(i,n) for(int i=1;i<=n;i++)
#define forrev(i,n) for(int i=n-1; i>=0; i--)
#define forab(i,a,b) for(int i=a;i<=b;i++)
#define forba(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 100005
#define inf 100000008
/*------------------------------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);
double u,v,w,s,x,y,z;
while((scanf("%lf %lf %lf",&u,&v,&w))==3)
{
if(u+v>w && v+w>u && u+w>v)
{
s=(u+v+w)/2.0;
x=(s*(s-u)*(s-v)*(s-w)*1.0);
y=sqrt(x)*1.0;
z=y*(4.0/3.0);
pf("%.3f\n",z);
}
else pf("-1.000\n");
}
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! 