UVA 10200 Prime Time【数论-素数区间判断+精度。前缀和】_euler isawell-knownmatematician, and, amongmanyoth-优快云博客

本文链接：https://blog.youkuaiyun.com/Li_Hongcheng/article/details/80101014

本文介绍了一种基于欧拉公式的素数生成方法，并提供了一个程序实现，该程序能够计算指定区间内由欧拉公式产生的素数比例。

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Euler is a well-known matematician, and, among many other things, he discovered that the formulan2 +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 knownthat 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 certaininterval.
Input
Each line of input will be given two positive integer a and b such that 0 ≤ a ≤ b ≤ 10000. You mustread 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 inthis 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<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=10005;
double eps =1e-8;
typedef long long ll;
int c[11111];

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

int main()
{
    for(int i=0;i<maxn;i++)
    {
        c[i]=c[i-1]+is_prime(i*i+i+41);
    }
    int a,b;
    while(~scanf("%d%d",&a,&b))
    {
        int d=c[b]-c[a-1];
        printf("%.2lf\n",d*100.0/(b-a+1)+eps);
    }

}