Prime Time UVA - 10200（打表）

最新推荐文章于 2020-03-02 12:32:40 发布

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最新推荐文章于 2020-03-02 12:32:40 发布

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该博客介绍了欧拉发现的公式n^2 + n + 41在0≤n<40时产生质数，虽然不总是产生质数，但在n≤10,000,000时仍有47.5%的情况产生质数。任务是编写程序计算给定区间内由公式产生的质数比例，并提供样例输入输出。" 116328372,10544093,11g ASM Quorum Failure Group 实验解析,"['数据库理论', 'Oracle ASM', '故障恢复']

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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

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<algorithm>

#define ll long long
#define inf 0x3f3f3f3f
#define maxn 1000007  // 1e6+7

using namespace std;
int num[100010];

int check(int n)
{
    for(int i=2;i*i<=n;i++)
    {
        if(n%i == 0)
            return 0;
    }
    return 1;

}
void init()
{
    for(int i=0;i<=10000;i++)
    {
        int tmp = i*i + i + 41;
        num[i] = num[i-1];
        if(check(tmp) == 1)
            num[i] ++;
    }
}
int main()
{
    init();
    int a,b;
    while(~scanf("%d%d",&a,&b))
    {
        int len = b-a+1;
        int tmp2 = num[b] - num[a];
        if(check(a*a+a+41) == 1)
            tmp2 ++;
        printf("%.2lf\n",( (tmp2*1.0) / (len*1.0) +1e-8 )* 100);    
    }
    return 0;
}