CALCULATING THE SIZE OF STRUCTURE IN C/C++:

consider a strucure

struct X
{
    short s; 
    int   i; 
    char  c; 
             
};

what do you think the size of structure will be? 2 + 4 + 1 ?

Sorry, then you are wrong. Because to reduce the lookup time by CPU compiler pads up the bytes untill it becomes a multiple of 4. This is however dependent on hardware,but most compilers does so.

so what happens is 

struct X
{
    short s; /* 2 bytes */
             /* 2 padding bytes */
    int   i; /* 4 bytes */
    char  c; /* 1 byte */
             /* 3 padding bytes */
};

but this is not the best way to declare the struct , why? because you waste space due to  padding of bytes. So here are some tricks to reduce the eat up of byte padding







struct Y
{
    int   i; /* 4 bytes */
    char  c; /* 1 byte */
             /* 1 padding byte */
    short s; /* 2 bytes */
};

struct Z
{
    int   i; /* 4 bytes */
    short s; /* 2 bytes */
    char  c; /* 1 byte */
             /* 1 padding byte */
};

POST YOUR DOUBTS IN THE COMMENTS

SPOJ : NEG2 SOLUTION

I suggest you read wikipedia link. But i can summarise you in a nut shell. Our task is to convert the given decimal to a negative base(2) number to binary form
i.e take for example  7(base 10)

Basically pick the remainder as the positive even when remainder is negative (tricky pay attention!)

 7 = -3*-2 + 1. (least significant digit)
-3 =  2*-2 + 1
 2 = -1*-2 + 0
-1 =  1*-2 + 1
 1 =  0*-2 + 1. (most significant digit)

as you can notice in 2nd step -3 % -2 is actually -1, but we have to store as 1 as binary representation cant contain negative numbers.

so we add 1 to n/2 i.e  -3/-2  = 1 adding 1 we get 2 ( this ensures remainder is always postive)


posting the logic:

  string s="";
     if(n==0)
     {
             printf("0\n");
             return;
     }
     int i=-2;

     while(n)
     {
             if(n%i<0)
             {

             s=(char)(((n%i)+2)+'0')+s;   // When the remainder is negative 

             n=(n/i)+1;                    // adding one so that when divided remainder is > 0
             }
             else
             {
             s=(char)((n%i)+'0')+s;
             n=n/i;
             }

     }
    cout<<s;

SPOJ : ACPC 10D Trigraphs Solution

This problem could be solved by using DP.
Step 1:
        Initialse the adjacent nodes to the source node.
Step 2:
        Loop Across the other i nodes from 2 to N and initialise them.

P re-Requesite: DP
     However Learning Floyd Warshall Algorithm would help you in solving this problem.



Code:


#include <iostream>
#include <string>
#include <stdio.h>
using namespace std;

int main()
{
int i=1;
while(1) {
int n;
cin >>n;
if(n==0)
return 0;
int a[n][3];
for (int i=0;i<n;i++)
cin >>a[i][0]>>a[i][1]>>a[i][2];
cout<<i<<". ";
if(n==1)
cout<<a[0][1]<<endl;
else {
                                                               // Initialisation Step
a[1][0] += a[0][1];
a[0][2] += a[0][1];
a[1][1] += min(min(a[1][0],a[0][1]),a[0][2]);
a[1][2] += min(min(a[0][2],a[0][1]),a[1][1]);


                                                                 // Calculating the remaing nodes
for(int i  = 2 ; i<n;i++)
{
a[i][0] += min(a[i-1][0],a[i-1][1]);
a[i][1] += min(min(min(a[i][0],a[i-1][0]),a[i-1][1]),a[i-1][2]);
a[i][2] += min(min(a[i][1],a[i-1][1]),a[i-1][2]);
}
cout <<a[n-1][1]<<endl;
i++;
}
}
return 0;
}