【UVa】10200 - Prime Time（打表）

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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

题意就是说用 n^2 + n +41 这个公式出来的数是素数的概率。

对0~10000打一个表就行了。

输出的时候注意四舍五入（这一点特别坑）

代码如下：

#include <stdio.h>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
#define f(x) (x*x+x+41)
int sum[10000+11];
bool isPrime(int x)
{
	int endd = sqrt(x);
	for (int i = 2 ; i <= endd ; i++)
	{
		if (x % i == 0)
			return false;
	}
	return true;
}
void getPrime()
{
	sum[0] = 1;
	for (int i = 1 ; i <= 10000 ; i++)
	{
		if (isPrime(f(i)))
			sum[i] = sum[i-1] + 1;
		else
			sum[i] = sum[i-1];
	}
}
int main()
{
	getPrime();
	int n;
	int l,r;
	double ans;
	while (scanf ("%d %d",&l,&r) != EOF)
	{
		if (l == 0)
			ans = (double)(sum[r]) / (r+1);
		else
			ans = (double)(sum[r] - sum[l-1]) / (r - l + 1);
		ans = (int)(ans*10000+0.5) / 100.0;
		printf ("%.2lf\n",ans);
	}
	return 0;
}