Prime Time（区间素数概率判定（高精度）//打表）

最新推荐文章于 2021-11-13 12:56:57 发布

原创最新推荐文章于 2021-11-13 12:56:57 发布 · 224 阅读

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探讨了著名数学家欧拉发现的一个公式，该公式在特定区间内能高效生成大量素数。博客深入分析了这一现象，并提供了一个程序示例，用于计算在给定区间内由欧拉公式产生的素数比例。

Euler is a well-known matematician, and, among many other things, he discovered that the formula n2 +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

最后不加1e-8就过不了。

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
int dis[10005];
int judge(int x)
{
	for(int i=2;i*i<=x;i++)
	{
		if(x%i==0)
		return 0;
	}
	return 1;
}
int main()
{
	for(int i=0;i<=10000;i++)
	{
		dis[i]=judge(i*i+i+41);
	}
	int n,m;
	while(~scanf("%d%d",&n,&m))
	{
		double sum=0;
		for(int i=n;i<=m;i++)
		sum+=dis[i];
		printf("%.2lf\n",(sum*1.0)/(m-n+1)*100+1e-8);
	}
}