1007 prime to conjecture (20 minutes)
Let d n is defined as: dn = p n + 1- p n, where p i is the i-th prime number. Clearly there is d 1 = 1, and for n> 1 there is an even number dn. "Primes to guess" that "there is an infinite number of adjacent and poor is a prime number 2."
Now given an arbitrary positive integer N (<105), does not exceed Calculate number N satisfies the conjecture of primes.
Input formats:
It gives a positive integer in the input line N.
Output formats:
In the output line does not exceed the number N satisfy the conjecture of the primes.
Sample input:
20
Sample output:
4
/ This question many ways, it is very simple, you can first save all primes less than N up to judge, side judge can traverse side (and I'm this way, to get three if the code is simple, take up less memory, this recommendation method) /
/ after writing the fifth point there was a timeout, internet search, the problem is when I judge a prime number, the cycle is from 2 to x / 2 (in fact, with 2 to sqrt (x) on They, sqrt (x) are repeated later) results in the timeout. /
/ Solution: When seeking prime, do not have from 2 to x-1 (x / 2 will timeout) to go cycling, only 2 to sqrt (x), otherwise for the last time out test points /
#include <stdio.h>
int ss(int x)
{
if(x==1 || x==2 || x==3 )
return 1;
for(int j=2;j<=sqrt(x);j++)
{
if(x%j==0)
return 0;
}
return 1;
}
int main()
{
int N,count=0;
scanf("%d",&N);
for(int i=1;i<=N-2;i++)
{
if(ss(i))
if(!ss(i+1))
if(ss(i+2))
count++;
}
printf("%d\n",count);
return 0;
}
/*这道题方法很多，也很简单，可以先把所有N以内的素数存起来判断，也可以边遍历边判断（我就是这种方法，三个if搞定代码简单，占用内存也少，推荐这个方法）*/
/*写完以后第五个点出现了超时的情况，网上搜了下，问题出在我判断素数的时候，循环过程是从2到x/2（其实用2-sqrt（x）就可以了，sqrt（x）后面的都是重复的）导致超时了。*/
/*解决方法：求素数时，不必从2到x-1(x/2也会超时)去循环，只需2到sqrt(x)即可，否则最后一测试点会超时*/
