URI Problem 1179 Solution Array Fill IV - URI Online Judge Solution

URI Online Judge Solution 1179 Array Fill IV Using C, Python Programming Language.

In this problem you need to read 15 numbers and must put them into two different arrays: par if the number is even or impar if this number is odd. But  the size of each of the two arrrays is only 5 positions. So every time you fill one of two arrays, you must print the entire array to be able to use it again for the next numbers that are read. At the end, all remaining numbers of each one of these two arrays must be printed beginning with the odd array. Each array can be filled how many times are necessary.

Input

The input contains 15 integer numbers.

Output

Print the output like the following example.

Input Sample	Output Sample
1 3 4 -4 2 3 8 2 5 -7 54 76 789 23 98	par[0] = 4 par[1] = -4 par[2] = 2 par[3] = 8 par[4] = 2 impar[0] = 1 impar[1] = 3 impar[2] = 3 impar[3] = 5 impar[4] = -7 impar[0] = 789 impar[1] = 23 par[0] = 54 par[1] = 76 par[2] = 98

Solution Using C:

#include <stdio.h>
int main()
{
    int a,b,c=0,d,e,f,g,h,i,j,l=0,m=0,x=0,n,p,Even[5],Odd[5],List[15];
    for(b=0; b<15; b++)
        scanf("%d", &List[b]);
    for(a=0; a<15; a++)
    {
        if(l==5)
        {
            for(n=0; n<5; n++)
                printf("impar[%d] = %d\n", n, Odd[n]);
            l=0;
        }
        if(m==5)
        {
            for(p=0; p<5; p++)
                printf("par[%d] = %d\n", p, Even[p]);
            m=0;
        }
        if(List[a]%2!=0)
        {
            Odd[c]=List[a];
            ++c;
            l++;
            if(c==5) c=0;
        }
        if(List[a]%2==0)
        {
            Even[x]=List[a];
            ++x;
            m++;
            if(x==5) x=0;
        }
    }
    for(i=0; i<l; i++)
            printf("impar[%d] = %d\n", i, Odd[i]);
    for(j=0; j<m; j++)
            printf("par[%d] = %d\n", j, Even[j]);
    return 0;
}