# Other UVA Problem Solutions

There are a lot of problems separated from **UVa** **Problem ** Categories like: **Ad hoc problem**,**Geometry** problem. These problem are known as Other** uva problem solutions**

Though these problem are included in other problems categories, these problem are more interesting ,easy & critical .I hope you will get pleasure by solving Other** uva problem solutions**.

other problems are easy and can be solved by previous knowledge. But be careful about time, space and special criteria when

solving them.

**Some Example of Other**** uva problem solutions**

**uva problem solutions**

**Problem A**

**Problem Name: Back To ****High School & College geometry :**

In above figure, ABCD is a **parallelogram** and AEB is a triangle where E lies on any point of CD. You are given x and y coordinates of A,B and C. You have to find out the area of triangle AEB.

**Input**

Each line of input to your program will contain the x and y coordinates of three points (A,B and C) in the order A_{x}A_{y}B_{x}B_{y}C_{x}C_{y}. These coordinates will be real (float or integer) numbers separated from each other by one or more spaces.

**Output:**

Print the area of triangle in one line with 2 digits after the decimal point. Print a new line after the end of each output. Be careful!!! Area is always a positive number.

**Sample input**

0 0 0 2 2 2

-1.5 -2.3 -3.1 4 2.9 3

0.1 9 2.5 -6 -4 -6.7

**Sample Output**

2.00

18.10

49.59

**Solution:**

It is very easy** geometry problem under Ad hoc problems categories .** Area of triangle AEB is equal to area of triangle ABC. And using co-ordinate geometry area of ABC is,

**Program Code:**

#include <cstdio> #include <cmath> using namespace std; int main() { float ax,ay,bx,by,cx,cy,d,x,y; while(scanf("%f %f %f %f %f %f",&ax,&ay,&bx,&by,&cx,&cy)==6) { printf("%.2f\n",fabs(0.5*(ax*(by-cy)+ay*(cx-bx)+(bx*cy-cx*by)))); } return 0; }

**Problem – No:2**

** uva Problem Name: Apple-The Innovation**

“Apple” the innovation of “Steve jobs”. He is best known as the co-founder, chairman, and chief executive officer of Apple Inc. Jobs has received a number of honors and public recognition for his influence in the technology and music industries. He has widely been referred to as “legendary”, a “futurist” or simply “visionary”, and has been described as the “Father of the Digital Revolution”, a “master of innovation”, and a “design perfectionist”. Unfortunately he died in October 5, 2011.

Once Eisty dreamed a dream that he met with Steve Jobs. Steve played a nice game with him. He told Eisty to take some imaginary apples & not to tell the number of apples. Eisty took 39 apples. Then Steve told to take same number of apples from his friend. Then told to take from him 12 apples and sum the total. Eisty did & then he had 90 apples. Steve told to give half of the total apples to the poor and return the apples of his friend. Then Steve told u have 6 apples remaining. Eisty was surprised to see how Steve could tell the remaining number as he didn’t know the number of apples !!

Steve is absent in the present world. But Eisty believes that God spread out the talents all around the world. That’s why he want to know how many Steve Jobs will be from all over the world and so want to play the same game that was played with Steve Jobs. Now your task is to do the same task as Steve did, input the number of apples like Steve provided and print the remaining number of apples.

**Input:
**Input starts with an integer T (T<=1000) , the number of test cases. Each test case will contain an integer. All integers will be less than 10,000 and greater than 0.You can assume that no operation overflows a 32-bit integer.

**Output:
**For each test case , output a single line giving the case number followed by the remaining numbers of apples. See the sample output for exact format.

Sample input: |
Sample Output: |

21011 |
Case 1: 5.00 Case 2: 5.50 |

**Solution:**

**Input File: To run this program you must save it as**: input.txt

20 // Test case

257

123

2569

124

5896

2548

12

10

11

25

24

2486

4

8

264

63

56

654

4862

96

**Program Code:**

#include<stdio.h> int main(){ int i,n; float a; freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); scanf("%d",&n); for(i=1;i<=n;i++){ scanf("%f",&a); printf("Case %d: %.2f\n",i,a/2); }

**Output File Looks Like Below and must save as it** : output.txt

Case 1: 128.50

Case 2: 61.50

Case 3: 1284.50

Case 4: 62.00

Case 5: 2948.00

Case 6: 1274.00

Case 7: 6.00

Case 8: 5.00

Case 9: 5.50

Case 10: 12.50

Case 11: 12.00

Case 12: 1243.00

Case 13: 2.00

Case 14: 4.00

Case 15: 132.00

Case 16: 31.50

Case 17: 28.00

Case 18: 327.00

Case 19: 2431.00

Case 20: 48.00