Prime Time

Prime Time链接

Euler is a well-known matematician, and, among many other things, he discovered that the formula
n
2 + n + 41 produces a prime for 0 ≤ n < 40. For n = 40, the formula produces 1681, which is 41 ∗ 41.
Even though this formula doesn’t always produce a prime, it still produces a lot of primes. It’s known
that for n ≤ 10000000, there are 47,5% of primes produced by the formula!
So, you’ll write a program that will output how many primes does the formula output for a certain
interval.
Input
Each line of input will be given two positive integer a and b such that 0 ≤ a ≤ b ≤ 10000. You must
read until the end of the file.
Output
For each pair a, b read, you must output the percentage of prime numbers produced by the formula in
this interval (a ≤ n ≤ b) rounded to two decimal digits.
Sample Input
0 39
0 40
39 40
Sample Output
100.00
97.56
50.00

大意

根据nn+n+41，来判断在一个范围内素数所占的比例，再100。注意精度，四舍五入。

#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int s[10011];
int prime(int l)
{
	int j;
	for(j=2;j<=(int)sqrt(l);j++)
	{
		if(l%j==0)
		return 0;
	}
	return 1;
}
void db()
{
	long long l,i,j;s[0]=1;
	for(int i=1;i<10011;i++)
	{
		l=i*i+i+41;
		if(prime(l))
		s[i]=s[i-1]+1;
		else
		s[i]=s[i-1];
	}
}
int main()
{
	int a,b,k;
	db();
	while(scanf("%d%d",&a,&b)!=EOF)
	{

		k=s[b]-s[a]+prime(a*a+a+41);
		double h=k*1.0/(b-a+1)*100;
		printf("%.2lf\n",h+1e-8);
	}
	return 0;
}