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Saturday, March 29, 2014

BLOPER - Operators

Link to the question

Solution of BLOPER spoj - spoj Operators

Maximum & Minimum Sum possible with N natural number is n*(n+1)/2 and out of it we need to take S. then we'll left with N - S. So, our aim is to make the final N-S effect zero therefore we divide N - S in two equal part and it only possible when N-S is an even number hence would be "Impossible" for odd count. After dividing if possible let us say S1,S2 (equal in magnitude), all we'll have to do is make S1 - S2.
Rest i leave it upto you. And think over it when S is negative.

#include<stdio.h>

int main()
{
    int arr[505]={0},i,j,n,s;

    scanf("%i%i",&n,&s);

    if(s < 0)
    {
        j = (n*(n+1))/2;

        if( (j+s)&1 || j <= -1*s )
        {
            printf("Impossible\n");
        }

        else
        {
            j = (j + s)/2 - 1;

            i = n+1;
            while(--i > 1 && j)
            {
                if( j - i > 1)
                {
                    j -= i;
                    arr[i] = 1;
                }

                else if(j-i == 0)
                {
                    j -= i;
                    arr[i] = 1;
                    break;
                }
            }

            if(j)
                printf("Impossible\n");
            else
            {
                printf("1");

                i = 1;
                while(++i <= n)
                {
                    arr[i] ? printf("+%i",i) : printf("-%i",i);
                }

                printf("\n");
            }

        }
    }

    else
    {
        j = (n*(n+1))/2;

        if( (j-s)&1 || j < s )
        {
            printf("Impossible\n");
        }

        else
        {
            j = (j - s)/2;

            i = n+1;
            while(--i > 1 && j)
            {
                if( j - i > 1)
                {
                    j -= i;
                    arr[i] = 1;
                }

                else if(j-i == 0)
                {
                    j -= i;
                    arr[i] = 1;
                    break;
                }
            }

            if(j)
            {
                printf("Impossible\n");
            }

            else
            {
                printf("1");

                i = 1;
                while(++i <= n)
                {
                    arr[i] ? printf("-%i",i) : printf("+%i",i);
                }

                printf("\n");
            }
        }
    }

    return 0;
}