SPOJ : KURUK 14 (Genie Sequence)

Hints: Consider the fact that for a sequence to be a genie sequence all the possible positions must be satisfied

Solution:
consider 6 elements 5 4 3 2 1 0
This forms a genie sequence as all the possible positions are covered respectively
But the position can be from either sides i.e the first number can be 5 or 0
                                                                       second number can be 4 or 1
                                                                       third number can be 3 or 2
and so on...
notice that sum of this combination ie always n - 1 here it is 6 - 1 = 5
(5+0) , (4+1),(3+2)... = 5

->  Thus create an array and input elements,intialise the array to 0
for each element mark in the arrray as 1 if it is already marked 1 , then mark the pair of that element..
for example if 5 is entered mark array[5] = 1 , but if 5 is entered again mark array[ n - 1 - 5] = 1

->after this process again iterate through the array, if u encounter an element with value 0 that means this cant form a genie sequence..


C code:
#include <stdio.h>
#include <stdlib.h>

int main(void) {
int t,n,temp,i,j,flag;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
int *a;
flag = 0;
a = calloc(n+1,sizeof(int));
for ( i = 0 ; i < n ; i++)
{
scanf("%d",&temp);
if(temp < n)
{
if(a[temp] == 0)
a[temp] = 1;
else
a[n -1 - temp] = 1;

}
}

for( i = 0 ; i<n;i++)
{

if(a[i] == 0)
{
flag  = 1;
break;
}
}
if(flag == 1)
printf("NO\n");
else
printf("YES\n");
}

return 0;

}

SPOJ : MANGOES Solution

Hints:
Just catch the point that GCD of 3 consecutive numbers must be 1.

Solution:
If GCD of three consecutive numbers is 1,the first number cant be an even  number because then the third number would obviously be an even number and so GCD wont be 1 

example
2,3,4
when first number if even (2) third number will be even(4) and GCD wont be 1 here it is 2.
So the first number must be an odd number ex:3,4,5 
Now comes the logic for three consecutive numbers only 2 can be a factor common to all three(if u cant figure out why feel free to comment or mail me,will reply ASAP)
so number of magoes = sum of numbers that satisfy the condition which in case is only the odd numbers
* Sum of odd numbers starting from 1 is n*n where n is the number of terms, here the number of terms is (N-1)/2
so number of mangoes is (N-1)/2*(N-1)/2
But we are required to give a number less than N so take modulo %

Here is the C code:

#include <stdio.h>

int main(void) {
long long int t,n,m;
scanf("%lld",&t);
while(t--)
{
scanf("%lld",&n);
m = (n -1)/2;
m = m*m;
m = m %n;
printf("%lld\n",m);
}
return 0;

}

SPOJ : ACT solution (Alpha Centauri Tennis)

Spoj:ACT question

Hints: 
The winner of the match is the winner of the last game

Solution:

#include<stdio.h>
#include<string.h>
#include<stdlib.h>


int main()
{
    int t,n;
    char *s;
    scanf("%d",&t);
    s = (char*)malloc(1000);

    while(t--)
    {
       
        scanf("%d %s",&n,s);
n=strlen(s)-1;
        printf("%c\n",s[n]);
    }
return 0;