UVA 10200 Prime Time （打表）(精度太坑了）

UVA 10200 Prime Time （打表）

Prime Time

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，打表倒不是问题，此题思路很清晰，就是精度卡的坑的很。最后不加1e-8就过不了，加了就过了。

若有幸哪位大牛知道原理的阔不阔以告诉我呀，评论一下就好

#include<cstdio>
#include<cmath>
using namespace std;
const int N = 10000+10;
int cnt[N]={1};
void init() {
	for(int i=1; i<N; i++) {
		int flag=0;
		int t=i*i+i+41;
		for(int j=2; j<=sqrt(t); j++) {
			if(t%j==0) {
				flag=1;
			}
		}
		if(!flag)
			cnt[i]=1;
	}
	for(int i=0;i<=10000;i++){
		cnt[i]=cnt[i]+cnt[i-1];
	}
}
int main() {
	int a,b;
	init();
	while(scanf("%d%d",&a,&b)!=EOF) {
		double s=cnt[b]-cnt[a-1];
		printf("%.2lf\n",s/(b-a+1)*100.0+1e-8);//不知道加个1e-8次方有什么用，不是让精确到小数点后两位吗
	}
	return 0;
}