Here on earth, our 24-hour day is composed of two parts, each of 12 hours. Each hour in each part has a corresponding hour in the other part separated by 12 hours: the hour essentially measures the duration since the start of the day part. For example, 1 hour in the first part of the day is equivalent to 13, which is 1 hour into the second part of the day.
Now, consider the equivalent hours that are both prime numbers. We have 3 such instances for a 24-hour 2-part day:
5~17
7~19
11~23
Accept two natural numbers D, P >1 corresponding respectively to number of hours per day and number of parts in a day separated by a space. D should be divisible by P, meaning that the number of hours per part (D/P) should be a natural number. Calculate the number of instances of equivalent prime hours. Output zero if there is no such instance. Note that we require each equivalent hour in each part in a day to be a prime number.
Example:
Input: 24 2
Output: 3 (We have 3 instances of equivalent prime hours: 5~17, 7~19 and 11~23.)
Constraints
10 <= D < 500
2 <= P < 50
Input
Single line consists of two space separated integers, D and P corresponding to number of hours per day and number of parts in a day respectively
Output
Output must be a single number, corresponding to the number of instances of equivalent prime number, as described above
Time Limit
1
Examples
Example 1
Input
36 3
Output
2
Explanation
In the given test case D = 36 and P = 3
Duration of each day part = 12
2~14~X
3~15~X
5~17~29 - instance of equivalent prime hours
7~19~31 - instance of equivalent prime hours
11~23~X
Hence the answers is 2.
Example 2
Input
49 7
Output
0
Explanation
Duration of each day part = 7
2~9~X~23~X~37~X
3~X~17~X~31~X~X
5~X~19~X~X~X~47
7~X~X~X~X~X~X
Hence there are no equivalent prime hours.
Answers
#include<stdio.h>
int count=0;
int prime(int a)
{
int flag=0;
for(int j=2; j<=(a/2);j++)
{
if(a%j==0)
{
flag=1;
break;
}
}
return flag;
}
int find(int b,int c,int d,int hrs)
{
int arr[10],ch=0;
for(int i=0;i<d;i++)
{
arr[i]=prime(b);
b=b+c;
if(b>hrs)
{
break;
}
}
for(int i=0;i<d;i++)
{
if(arr[i]==0)
ch++;
}
if(ch==d)
return 1;
else
return 0;
}
int main()
{
int D;
int P;
scanf("%d",&D);
scanf("%d",&P);
if(D%P==0)
{
int q=D/P;
for(int i=2;i<=D;i++)
{
count=count+find(i,q,P,D);
}
printf("%d",count);
}
return 0;
}