SAMER08F - Feynman

Problem is available at http://www.spoj.com/problems/SAMER08F/

This problem is easy to solve if you understand the pattern.

If n = 1, there is one 1-by-1 square.

If n = 2, there is one 2-by-2 square and four 1-by-1 squares.

If n = 3, there is one 3-by-3 square, four 2-by-2 squares and nine 1-by-1 squares.

so if n=k , there is one k^2 + (k-1)^2 + (k-2)^2 + (k-3)^2 + ....... + 1^2 squares.


code is given below in java :


import java.util.Scanner;
import java.lang.*;

class spoj 
{


public static void main(String args[])
{

Scanner s=new Scanner(System.in);
int val=0;
int sum=0;
int i=0;

while((val=s.nextInt())!=0){


for(i=1;i<=val;i++){

sum=sum+(int)(Math.pow(i,2));

}

System.out.println(sum);


sum=0;

}

}

}

NSTEPS-Number Steps

The problem is available at http://www.spoj.com/problems/NSTEPS/

The problem is very easy to solve.

solution is given below in java :




import java.util.Scanner;
import java.lang.*;

class spoj
{


public static void main(String args[])
{

Scanner s=new Scanner(System.in);
int count=s.nextInt();
int i=0;
int valx=0;
int valy=0;


while(i<count){

valx=s.nextInt();
valy=s.nextInt();

if(valx==valy){
if (valx % 2 == 0)
System.out.println(valx +valy);
else
System.out.println(valx +valx - 1);
}
else if(valy==(valx-2)){
if (valx % 2 == 0)
System.out.println(valx + valy);
else
System.out.println(valx + valy - 1);
}
else{
System.out.println("No Number");
}

i++;

}




}
}