区间素数概率判定（高精度）

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

精度太坑

#include<cstdio>
int su[10005];
int judge(int n)
{
    int i;
    for(i = 2 ; i*i <= n ; i++)
    {
        if(n % i == 0)
        return 0;
    }
    return 1;
}
int main()
{
    int i;
    for(i = 0 ; i <= 10005 ; i++)
    {
        su[i]=judge(i*i+i+41);
    }
    int a,b;
    while(scanf("%d %d",&a,&b)!=EOF)
    {
        double sum=0;
        for(i = a ; i <= b ; i++)
        {
            sum+=su[i];
        }
        printf("%.2f
",sum*1.0/(b-a+1)*100);
    }
}