Binary Search Eqn Solve Added

Author: zarif98sjs

Algorithm/16 Binary Search (Eqn Solve).cpp

@@ -0,0 +1,92 @@
+
+/**
+
+Binary Search (Eqn Solve)
+=========================
+
+Demo Problem : LightOj 1043 ( Triangle Partitioning )
+------------------------------------------------------
+
+        A
+        /\
+       /  \
+     D/----\E
+     /      \
+    /        \
+   B----------C
+
+You are given AB, AC and BC. DE is parallel to BC.
+You are also given the area ratio between ADE and BDEC. You have to find the value of AD.
+
+Idea
+----
+
+AB/AD = AC/AE = BC/DE
+
+So  , AE = (AC*AD)/AB
+and , DE = (BC*AD)/AB
+
+The area ratio of those two triangles can be calculated as a function of AD .
+So, we can Binary Search over AD
+
+**/
+
+/** Which of the favors of your Lord will you deny ?* */
+
+#include<bits/stdc++.h>
+using namespace std;
+
+double triangleRatio (double ab,double ac,double bc,double ad)
+{
+    double ae=(ac*ad)/ab, de=(bc*ad)/ab,s1,s2;
+
+    s1 = (ab+ac+bc)/2.0;
+    s2 = (ad+ae+de)/2.0;
+
+    double areaBig = sqrt(s1 * (s1-ab) * (s1-ac) * (s1-bc));
+    double areaSmall = sqrt(s2 * (s2-ad) * (s2-ae) * (s2-de));
+
+    double areaTrap = areaBig-areaSmall;
+
+    double ratio = areaSmall/areaTrap;
+
+    return ratio;
+
+}
+
+double EPS = 0.00000001;
+
+double BS(double lo,double hi ,double ab,double ac,double bc,double target)
+{
+    double mid;
+
+    do
+    {
+        mid = (hi + lo) / 2.0;
+
+        double value = triangleRatio(ab,ac,bc,mid);
+
+        if (value>target) hi = mid;
+        else lo = mid;
+    }
+    while ((hi - lo) > EPS);
+
+    return mid;
+}
+
+int main()
+{
+    int n;
+    scanf("%d",&n);
+
+    for(int i=1; i<=n; i++)
+    {
+        double ab,ac,bc,ratio;
+        scanf("%lf %lf %lf %lf",&ab,&bc,&ac,&ratio);
+
+        printf("Case %d: %lf\n",i,BS(0,ab,ab,bc,ac,ratio));
+
+    }
+
+    return 0;
+}