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Jishan's Log 14: Codeforces 69A solution

Codeforces 69A: Young Physicist The Problem: A boy is given a boy with a set of forces acting on it in x, y and z space. He needs to find out if the body is in equilibrium or not. For a body to be in equilibrium, all forces on it must be balanced ( Σ(F) == 0 ). So, we need to find out if the forces in x, y, z plane individually are 0. The Solution: The Algorithm: 1) Take n input. Declare sums of x, y, z and set them to 0. 2) Take a loop and make it run n times. 3) Take the x, y, z inputs and sum them up with their respective sum variables. 4) Check to see if all of them are zero. If true, print YES. Else print NO The Code: #include <bits/stdc++.h> #define   fastio   ios :: sync_with_stdio ( 0 );  cin . tie ( 0 );  cout . tie ( 0 ); using   namespace   std ; int   main () {      fastio ;      int   n ;  cin   >>   n ;      int   x_sum ,  y_sum ,...
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Jishan's Log 6: Codeforces 158A solution

 Codeforces 158A: Next Round What the problem wants? The problem states that there are a n scores and from those scores we must take the score of the kth position as the passing mark. If a score is greater than or equal to the kth position and if it is greater than 0, then the candidate passes.  How to solve it? Algorithm: 1) take the n and k inputs 2) for i=0 => n (1 => n), we read the integers and store it in an array. We then get the value of the kth position and check if the value in the array is greater than k or not & if it is greater than 0. If it is, we increment a counter. Finally we print it. The Code: ***NOTE: Since I've taken the array from 0 => n, that's why I've taken the k-1 position. If you start from 1, you can take k only #include <bits/stdc++.h> using   namespace   std ; #define   fastio   ios :: sync_with_stdio ( 0 );  cin . tie ( 0 );  cout . tie ( 0 ); int   main () {      fastio ; ...
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Jishan's Log 8: Codeforces 263A solution

Codeforces 263A: A Beautiful Matrix What the problem wants? The problem states that a 5x5 matrix is given. It is full of 0's except one position has the number 1. We need to find the minimum moves needed to shift 1 from its current position to the center of the matrix. How to solve it? The Algorithm: 1) take a matrix and its input. 2) search for 1. If found, then find the absolute value from where it is to the center of the matrix. Do this for both i & j position.  ***NOTE: The logic here is that the minimum moves made will only be through the horizontal or vertical movement. not cross movements. So we take the i at that position and subtract it to the i of the center element. Same for j. The Code: #include <bits/stdc++.h> using   namespace   std ; #define   fastio   ios :: sync_with_stdio ( 0 );  cin . tie ( 0 );  cout . tie ( 0 ); void   minMoves ( int   matrix [ 5 ][ 5 ]) {      int   i_idx   = ...
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